§11 · Agent X1 · Symplectic counts · 2026-10-17T20:40:47Z

Correction X1.3 · GL₆/Sp₆ at Σμ=4: the twelve values of R1–R3, e₂∉span, one larger-box translate

proved correction

Correction. Corrects Dead end 4.10 · $\GL_6/\Sp_6$ at the label $(1,1,1,1,0,0)$: the three in-box translates do not suffice; a larger box is needed.

The twelve values of the in-box translates $R_1,R_2,R_3$ of Dead end 4.10 on $c_1,\dots,c_4$ are proved as polynomials in $q$ by explicit counts, and confirmed by direct lattice enumeration at $q=2,3,5$; four of Dead end 4.10's values are false, $e_2\notin\operatorname{span}(R_1,R_2,R_3)$ still holds, and $T_{(2,0,0,0,0,0)}\ast\varphi_0=q(q+1)\mathbf 1_{c_1}$ completes the span.

Correction X1.3.

Let $X=\mathrm{GL}_6/\mathrm{Sp}_6$ over any non-archimedean $F$ with residue field $\mathbb F_q$. Let $c_1,\dots,c_4$ be the labels $(2,2,0,0,0,0)$, $(1,1,1,1,0,0)$, $(2,2,1,1,-1,-1)$, $(2,2,2,2,-2,-2)$, represented by $x=\bigoplus_pt^{a_p}J_2$ with $a=(2,0,0),(1,1,0),(2,1,-1),(2,2,-2)$, and let $R_i=T_{\lambda_i}\ast\varphi_0$ with $\lambda_1=(1,1,0,0,0,0)$, $\lambda_2=(1,1,1,0,0,-1)$, $\lambda_3=(1,1,1,1,-1,-1)$ (lattice types $-w_0\lambda_i$: $(0,0,0,0,-1,-1)$, $(1,0,0,-1,-1,-1)$, $(1,1,-1,-1,-1,-1)$). Then, for every $q$, the $R_i$ vanish off $c_1,\dots,c_4$ and

Table 2. and $c_1$ and $c_2$
	$c_1$	$c_2$	$c_3$	$c_4$
$R_1$	$1$	$(q+1)(q^2+1)$	$0$	$0$
$R_2$	$q(q+1)(q^2+1)$	$q^3(q+1)^2(q^2+1)$	$(q+1)^2$	$0$
$R_3$	$q^3(q+1)(q^2+1)$	$0$	$q(q^2-1)$	$1$

Dead end 4.10's values of $R_2(c_1)$, $R_2(c_2)$, $R_3(c_1)$, $R_3(c_3)$ are false; its other eight values are right. (This confirms Correction 10.7, part 1.)
$e_2:=\mathbf 1_{\mathcal O_{c_2}}\notin\operatorname{span}(R_1,R_2,R_3)$: confirmed, but the elimination must use the $c_4$-entry first, since $R_3(c_3)\ne0$.
$T_{(2,0,0,0,0,0)}\ast\varphi_0=q(q+1)\mathbf 1_{\mathcal O_{c_1}}$ (lattice type $(0,0,0,0,0,-2)$, outside the box), so

$$\mathbf 1_{\mathcal O_{(1,1,1,1,0,0)}}=\frac1{(q+1)(q^2+1)}\Big(T_{(1,1,0,0,0,0)}-\frac1{q(q+1)}T_{(2,0,0,0,0,0)}\Big)\ast\varphi_0\in\mathcal H_K\varphi_0 :$$

one translate from the box $t^2\mathcal O^6\subseteq M\subseteq t^{-2}\mathcal O^6$ completes the span, exactly as $T_{(2,0,0,0)}$ does for $\mathrm{GL}_4/\mathrm{Sp}_4$ (Correction X1.1).

Counting problem. As in Correction X1.1, step 1, $R_i(x)$ is the number of lattices $M$ with $\operatorname{inv}(\mathcal O^6,M)=-w_0\lambda_i$ and $M=M^*$, $M^*=\{v:x(v,M)\subseteq\mathcal O\}$. The three types are the types with entries in $[-1,1]$ and sum $-2$, i.e. $t\mathcal O^6\subseteq M\subseteq t^{-1}\mathcal O^6$.

Support and reduction. Integrality on $t\mathcal O^6$ gives $a_p\ge-2$; $M=M^*\supseteq(t^{-1}\mathcal O^6)^*=\bigoplus_pt^{1-a_p}\mathcal O^2$ together with $M\subseteq t^{-1}\mathcal O^6$ gives $a_p\le2$; the determinant gives $\sum a_p=2$. The labels with $a_1\ge a_2\ge a_3$ in $[-2,2]$ and sum $2$ are exactly $c_1,\dots,c_4$. With $A:=t\mathcal O^6+(t^{-1}\mathcal O^6)^*$, every such $M$ satisfies $A\subseteq M\subseteq A^*$, and $M\mapsto L=M/A$ is a bijection onto the Lagrangian $\mathcal O$-submodules of $Q=A^*/A$ (the reduction of Correction 10.7, step 1, re-derived).

$c_4$. $A=t^{-1}\mathcal O^4\oplus t\mathcal O^2=A^*$: one lattice, of type $(1,1,-1,-1,-1,-1)$. So $R_3(c_4)=1$, $R_1(c_4)=R_2(c_4)=0$.

$c_3$. $A=t^{-1}\mathcal O^2\oplus\mathcal O^2\oplus t\mathcal O^2$, $A^*=t^{-1}\mathcal O^4\oplus\mathcal O^2$, $Q=H_1\perp H_2$ with $H_1=t^{-1}\mathcal O^2/\mathcal O^2$, $H_2=\mathcal O^2/t\mathcal O^2$ symplectic planes over $\mathbb F_q$. Block 1 contributes $(-1,-1)$; in coordinates $3,\dots,6$ the type is $(1^r,0^{4-2r},(-1)^r)$ with $r=\dim\mathrm{pr}_{H_1}L=2-\dim(L\cap H_2)$. $r=0$ would make $L=H_2$, not isotropic. $r=1$: $L=\ell\oplus\ell'$ with $\ell\subset H_2$, $\ell'\subset H_1$ lines: $(q+1)^2$, type of $R_2$. $r=2$: the remaining Lagrangian planes, $(q+1)(q^2+1)-(q+1)^2=q(q^2-1)$, type of $R_3$. So $c_3\mapsto(0,(q+1)^2,q(q^2-1))$.

$c_1$. $A=t^{-1}\mathcal O^2\oplus t\mathcal O^4$, $A^*=t^{-1}\mathcal O^6$; so $M=t^{-1}\mathcal O^2\oplus M'$ with $L'=M'/t\mathcal O^4$ a Lagrangian submodule of $R_2^4$ ($R_2=\mathcal O/t^2$) for the perfect form $t^{-2}J_4$. The type of $M'$ is $(0,0,0,0)$, $(1,0,0,-1)$, $(1,1,-1,-1)$ according as $L'$ has module type $(1^4)$, $(2,1,1)$, $(2,2)$. These were counted in Correction X1.2, step 4 (column $(2,2,0)$, the same module): $1$; $q(q+1)(q^2+1)$ ($L'=\mathcal Ou+t\widetilde{\bar u^\perp}$); $q^3(q+1)(q^2+1)$ ($R_2$-points of the smooth $\mathrm{LG}(2,4)$). Correction 10.7 obtains the last two from Macdonald's count for $\mathrm{Sp}_4$; the derivations agree.

$c_2$ (orthogonal-sum count, independent of Correction 10.7's $\ell^\perp/\ell$ reduction). $A=\mathcal O^4\oplus t\mathcal O^2$, $A^*=t^{-1}\mathcal O^6$, $Q=V\perp W$ with $V=t^{-1}\mathcal O^4/\mathcal O^4\cong\mathbb F_q^4$ (form $t^{-1}J_4$) and $W=t^{-1}\mathcal O^2/t\mathcal O^2\cong R_2^2$ (form $t^{-2}J_2$). The image of $M$ in $t^{-1}\mathcal O^6/\mathcal O^6$ is $L/(L\cap tW)$, so the type has $4-d$ entries $-1$ and, by $\sum=-2$, $2-d$ entries $+1$, where $d=\dim(L\cap tW)$: $d=2,1,0$ give $R_1,R_2,R_3$. For a Lagrangian $L$ of $V\perp W$, put $K_V=L\cap V$, $K_W=L\cap W$; comparing $|L|=|K_V||\mathrm{pr}_WL|=|K_W||\mathrm{pr}_VL|$ with $|L|^2=|V||W|$ and $\mathrm{pr}_VL\subseteq K_V^\perp$, $\mathrm{pr}_WL\subseteq K_W^\perp$ shows that $L$ is the graph of an anti-isometry $K_V^\perp/K_V\to K_W^\perp/K_W$, and conversely. The isotropic submodules of $W$ are $0$, the $q+1$ lines $\ell\subset tW$, the $q(q+1)$ cyclic $\mathcal Ow$ ($w$ primitive), and $tW$.
- $K_W=0$: $W$ is not killed by $t$, $K_V^\perp/K_V$ is; impossible.
- $K_W=\ell$: $K_W^\perp=\mathcal Ow_0+tW$, $K_W^\perp/K_W\cong\mathbb F_q^2$ symplectic; $K_V$ a line ($(q+1)(q^2+1)$), $q(q^2-1)$ anti-isometries; $d=1$.
- $K_W=\mathcal Ow$: Lagrangian, so $K_V$ is Lagrangian ($(q+1)(q^2+1)$); $d=\dim t\mathcal Ow=1$.
- $K_W=tW$: $K_V$ Lagrangian; $d=2$. Hence $R_1(c_2)=(q+1)(q^2+1)$, $R_2(c_2)=(q+1)(q^2+1)\big[(q+1)q(q^2-1)+q(q+1)\big]=q^3(q+1)^2(q^2+1)$, $R_3(c_2)=0$.

Comparison with Dead end 4.10. It had $R_2(c_1)=(q^4-1)/(q-1)$, $R_2(c_2)=(q+1)(q^2+1)(q^2+q)$, $R_3(c_1)=(q+1)(q^2+1)$, $R_3(c_3)=0$; at $q=3$: $40,480,40,0$ against the true $120,4320,1080,24$. The other eight entries agree.

Statement 2. If $e_2=\alpha R_1+\beta R_2+\gamma R_3$: the $c_4$-entry gives $\gamma=0$; then the $c_3$-entry gives $(q+1)^2\beta=0$; then the $c_1$-entry gives $\alpha=0$; the $c_2$-entry gives $0=1$. Since the in-box translates of degree $-2$ are exactly $R_1,R_2,R_3$ and $C_K$ is graded by $\operatorname{val}\operatorname{Pf}$, no combination of in-box translates equals $e_2$.

Statement 3. By Lemma 1 of Lemma 5.6 (re-proved in Correction X1.2, step 1) with $d=2$, $T_{(2,0,0,0,0,0)}\ast\varphi_0=q(q+1)\mathbf 1_{\mathcal O_{(2,2,0,0,0,0)}}$, and $R_1=\mathbf 1_{\mathcal O_{c_1}}+(q+1)(q^2+1)e_2$.

Machine check (different method from Correction 10.7's enumeration of Lagrangians of $Q$). Direct Hermite-normal-form enumeration of all $x$-unimodular $M\subseteq t^{-1}\mathcal O^6$ over $\mathbb F_q[\![t]\!]$, types by Smith reduction: all twelve entries agree with the table for $q=2,3,5$ (at $q=5$: $R_1=(1,156,0,0)$, $R_2=(780,117000,36,0)$, $R_3=(19500,0,120,1)$); $T_{(2,0,\dots,0)}$ gives $(6,0,0,0)$ and $(12,0,0,0)$ at $q=2,3$; for seven labels with $\sum a=2$ and an entry outside $[-2,2]$ there is no in-box unimodular lattice ($q=2,3$). The counts in steps 3–6 use only configurations over $\mathbb F_q$ and $\mathcal O/t^2$ and hold for every $F$ and $q$.

Builds on

Dead end 4.10 · $\GL_6/\Sp_6$ at the label $(1,1,1,1,0,0)$: the three in-box translates do not suffice; a larger box is needed
Theorem 4.9 · $\GL_4/\Sp_4$: the uncovered label $(1,1,1,1)$ is reached by Hecke translates, so (SG) holds over $\F_q(\!(t)\!)$ and $\Csat=\cC_K$
Correction 10.7 · Correction: the in-box Hecke translates of $\GL_6/\Sp_6$ at $\sum\mu=4$, and one translate that reaches $(1,1,1,1,0,0)$
Correction X1.2 · GL₆/Sp₆: Lemmas 1–3 re-proved; the |b|=4 table proved for all q; |b|≤5 computed

Built on by

Re-derivation X1.5 · (SG) for GL₂ₘ/Sp₂ₘ re-derived from Lemmas 2–3 alone; A4-13's open cells now decided
Re-derivation X1.6 · The verdict inputs downstream of A1-4, A4-10, A5-6, A7-11: what changes
Re-check V1 · Re-check by V1
Summary 9.6 · Verdict after the repairs: every input held or repaired; $E_6/F_4$ still the one divergence

Sources

/private/tmp/claude-501/-Users-miltonlin-Projects-Trees/f0f18b89-aa0d-4ce6-93d2-78c269eb7d7d/scratchpad/salvage/S1/engine.py
/private/tmp/claude-501/-Users-miltonlin-Projects-Trees/f0f18b89-aa0d-4ce6-93d2-78c269eb7d7d/scratchpad/salvage/S1/gl6_box.py
/private/tmp/claude-501/-Users-miltonlin-Projects-Trees/f0f18b89-aa0d-4ce6-93d2-78c269eb7d7d/scratchpad/salvage/S1/gl6_box_q.py
/private/tmp/claude-501/-Users-miltonlin-Projects-Trees/f0f18b89-aa0d-4ce6-93d2-78c269eb7d7d/scratchpad/salvage/S1/gl6_extra.py

	\(c_1\)	\(c_2\)	\(c_3\)	\(c_4\)
\(R_1\)	\(1\)	\((q+1)(q^2+1)\)	\(0\)	\(0\)
\(R_2\)	\(q(q+1)(q^2+1)\)	\(q^3(q+1)^2(q^2+1)\)	\((q+1)^2\)	\(0\)
\(R_3\)	\(q^3(q+1)(q^2+1)\)	\(0\)	\(q(q^2-1)\)	\(1\)

Table 1. Page production by session
	tokens	US $
This conversation: the search, the audit, the pages and propositions, and the VI characteristic-p run01:44–05:19 UTC	802 M	$490
Setup pages16:17–19:42 UTC	37 M	$19
Shareable animation17:16–17:22 UTC	1 M	$1
Proposition IIb: figures and the linear-algebra lemma19:48–20:05 UTC	14 M	$7
Audit pass: the repairs by Agents X1–X4, their re-checks and Propositions VI–VII (with the audit skill)19:35–02:12 UTC	275 M	$113
Proposition IX: converse search, verdict and trace formalism19:56–02:12 UTC	199 M	$106
Audit pass on the VI characteristic-p run; the simplified VIb page and the VI corrections05:21–07:33 UTC	268 M	$107
Removing the detail switch; Proposition IX restated; the hub and verdict reviewed and simplified05:34–15:56 UTC	38 M	$20
C_K for the categorical statements, in the Iwahori run's conversation18:27–18:32 UTC	9 M	$3
Total	1,642 M	$866