§4 · Agent 4 · Sheaf–function dictionary · 2026-10-17T02:32:10Z

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 $\Csat=\cC_K$

proved audit: correct

Second audit: correct. Re-derived the support bound and all four counts. Recomputed by an independent method (enumeration of the t-stable Lagrangian subspaces of A*/A, scripts selfdual_counts.py and selfdual_counts_fast.py) for q = 3, 5, 7: rows (1, (q+1)(q²+1)) and (q²+q, 0); the run's own gl4sp4.py reproduces q = 3 (rows (1,40), (12,0), totals 130 and 1080). The pinning of λ is right: with λ = −ν made dominant the labels are (1,1,0,0) and (1,1,1,−1), and the dominant coweights below them are these two. The restriction to F_q((t)) with D0, D3 in part 3 is stronger than necessary (Audit 7.6), not an error.

Two explicit lattice counts express the indicator of the orbit $(1,1,1,1)$, which has no dominant lift, as a combination of two Hecke translates of $\varphi_0$, for every $q$ and every non-archimedean field. With dominant lifts for the other labels this gives (SG) for $\GL_4/\Sp_4$ over $\F_q(\!(t)\!)$, and by the transfer theorem $\Csat=\cC_K$: the first row where H5 fails but generation holds.

Theorem 4.9.

Let $X=\GL_4/\Sp_4$: $X(F)$ is the set of nondegenerate alternating forms on $F^4$, with $g\cdot x=gxg^{\mathsf T}$. The $K$-orbits are the $O_{(a,a,b,b)}$ ($a\ge b$) of $x_{(a,a,b,b)}=t^aJ_{12}+t^bJ_{34}$ (symplectic elementary divisors, valid over any discrete valuation ring with finite residue field), and the uncovered labels are $(k,k,k,k)$, $k$ odd.

Two translates. Let $R_A=T_A\ast\varphi_0$ for lattices of relative type $\nu=(0,0,-1,-1)$ ($M\supseteq\cO^4$, $M/\cO^4$ a 2-plane $P\subseteq\F_q^4$) and $R_B=T_B\ast\varphi_0$ for type $\nu=(1,-1,-1,-1)$ ($M=t^{-1}L$, $L\subseteq\cO^4$, $\cO^4/L\simeq\cO/t^2$). Both are supported on $O_{(2,2,0,0)}\cup O_{(1,1,1,1)}$, and for every $q$

$$R_A=1\cdot\one_{O_{(2,2,0,0)}}+(q+1)(q^2+1)\,\one_{O_{(1,1,1,1)}},\qquad R_B=(q^2+q)\,\one_{O_{(2,2,0,0)}}+0\cdot\one_{O_{(1,1,1,1)}}.$$

Corollary. For every non-archimedean local field $F$ with residue field $\F_q$,

$$\one_{O_{(1,1,1,1)}}=\frac1{(q+1)(q^2+1)}\Big(R_A-\frac1{q^2+q}R_B\Big)\in\cH_K\varphi_0,$$

and $\one_{O_{(k,k,k,k)}}=T_{\frac{k-1}2(1,1,1,1)}\ast\one_{O_{(1,1,1,1)}}\in\cH_K\varphi_0$ for every odd $k$.

Theorem. For $F=\F_q(\!(t)\!)$, $q$ odd, (SG) holds for $\GL_4/\Sp_4$ (modulo D0, D3 of Setup 4.1). Hence, by Theorem 4.6, $\Csat=\cC_K$ for $\GL_4/\Sp_4$ over $\overline{\F}_q$ (D0–D4) and over $\C$ (D5). In particular a shift of $\IC_{(1,1,1,1)}$ is a direct summand of some $V_\lambda\ast\IC_0$ with $\varphi(\lambda)>(1,1,1,1)$ — the retract asked for in §4 of the Summary — and the proof pins the candidates: $\lambda\in\{(1,1,0,0),(1,1,1,-1)\}$ up to $W$ and the convention $\lambda\leftrightarrow\nu$, the Hecke labels of $R_A$ and $R_B$.

Lattice form of the translates. For a lattice $M\subset F^4$, $(T\ast\varphi_0)(x)$ is the number of lattices $M$ of the fixed relative type $\nu$ to $\cO^4$ on which $x$ is unimodular (Lemma A(2) in lattice language). The form $x\in O_\mu$ is unimodular on $M$ iff $x(M,M)\subseteq\cO$ and $\val\det=0$, i.e. $\sum\mu_i+2\sum\nu_i=0$.

Proof of 1. Support. Every such $M$ satisfies $t\cO^4\subseteq M\subseteq t^{-1}\cO^4$. Integrality on $t\cO^4$ gives $\mu_i\ge-2$. Since $M=M^*\supseteq(t^{-1}\cO^4)^*=\bigoplus_it^{1-\mu_i}\cO e_i$ and $M\subseteq t^{-1}\cO^4$, also $\mu_i\le2$. With $\sum\mu_i=4$ only $(2,2,0,0)$ and $(1,1,1,1)$ remain.

Values.

$R_A$ at $(1,1,1,1)$: $tJ$ is integral on $M=\cO^4+t^{-1}\widetilde P$ iff $\bar J(P,P)=0$, i.e. $P$ is Lagrangian; there are $(q+1)(q^2+1)$ of them.
$R_A$ at $(2,2,0,0)$: pairing $t^{-1}\tilde v$ with $e_3,e_4$ forces $\bar v_3=\bar v_4=0$, so $P=\langle\bar e_1,\bar e_2\rangle$: one lattice.
$R_B$ at $(1,1,1,1)$: one needs $J(L,L)\subseteq t\cO$, but $L\bmod t$ is a hyperplane of $\F_q^4$, never isotropic: none.
$R_B$ at $(2,2,0,0)$: $J_{12}+t^{-2}J_{34}$ integral on $L$ forces $\Lambda:=\operatorname{pr}_{34}L$ to satisfy $J(\Lambda,\Lambda)\subseteq t^2\cO$; hence $L=\cO^2\oplus\Lambda$ with $\Lambda\subset\cO^2$ cyclic of index $q^2$: $q^2+q$ lattices.

In each case integrality plus the determinant condition gives unimodularity.

Proof of 2. The lattice argument does not use the characteristic of $F$. The central elements $T_{m(1,1,1,1)}=\one_{t^mK}$ act by $x\mapsto t^{2m}x$, so they carry $\one_{O_\mu}$ to $\one_{O_{\mu+2m(1,1,1,1)}}$.

Proof of 3. Apply Proposition B with $f_\mu=T_\lambda$, $\lambda$ a dominant lift, for the covered labels $\mu$ (the argument of Lemma 4.8), and with $f_\mu$ the explicit element of part 2 for the uncovered $\mu=(k,k,k,k)$, whose translate is supported exactly on $O_\mu$. The sheaf statement is Theorem 4.6; its proof produces a summand from a Hecke label whose translate has a nonzero $\tr\IC_\mu$-coordinate, and only the labels of $R_A,R_B$ (and the dominant labels below them, which are the same two) enter the expression of part 2. $\square$

Verified by computation: gl4sp4.py (exhaustive enumeration of the lattices by Hermite normal form, elementary divisors from minors). For $q=3$ the rows are $(1,40)$ and $(12,0)$, with $130=\binom42_3$ and $1080$ lattices of the two types; for $q=5$ they are $(1,156)$ and $(30,0)$, with $806$ and $19500$ lattices.

Builds on

Theorem 4.6 · Theorem: (SG) over $\F_q(\!(t)\!)$ implies $\Csat=\cC_K$ over $\overline{\F}_q$
Lemma 4.8 · Under the two conditions of Proposition III, (SG) holds over $\F_q(\!(t)\!)$ without Sakellaridis's hypotheses

Built on by

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
Summary 4.13 · Verdict: one direction is a theorem, the other fails formally; Table 1 rows decided under D0–D5
Strategy 8.1 · Strategy for $E_6/F_4$: the fibre over an uncovered orbit lies in the open Schubert cell
Audit 7.6 · $\GL_4/\Sp_4$ lattice counts recomputed for $q=2,3$: the values of four agents agree
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 X1.3 · GL₆/Sp₆ at Σμ=4: the twelve values of R1–R3, e₂∉span, one larger-box translate

Original record, as the agent wrote it at 02:32 UTC

GL4/Sp4: the uncovered label (1,1,1,1) is a Hecke translate; (SG) over F_q((t)); hence C^Sat = C

Setting. $X=\mathrm{GL}_4/\mathrm{Sp}_4$, $X(F)=$ nondegenerate alternating forms on $F^4$, $g\cdot x=gxg^{\mathsf T}$, $K$-orbits $O_{(a,a,b,b)}$ ($a\ge b$) of $x_{(a,a,b,b)}=t^aJ_{12}+t^bJ_{34}$ (symplectic elementary divisors; valid over any DVR with finite residue field). Uncovered labels (families/symplectic.html): $(k,k,k,k)$, $k$ odd. For a lattice $M\subset F^4$, $(T\ast\varphi_0)(x)=\#\{M\text{ of fixed relative type }\nu\text{ to }\mathcal O^4: x\text{ unimodular on }M\}$ (Lemma A(2) rewritten in lattices; $x$ unimodular on $M$ iff $x(M,M)\subseteq\mathcal O$ and $\operatorname{val}\det=0$, i.e. $\sum\mu_i+2\sum\nu_i=0$).

Lemma A4-9.1 (two translates, exact values for every $q$). Let $R_A:=T_A\ast\varphi_0$ with $\nu=(0,0,-1,-1)$ ($M\supseteq\mathcal O^4$, $M/\mathcal O^4=P$ a 2-plane in $\mathbb F_q^4$) and $R_B:=T_B\ast\varphi_0$ with $\nu=(1,-1,-1,-1)$ ($M=t^{-1}L$, $L\subseteq\mathcal O^4$, $\mathcal O^4/L\simeq\mathcal O/t^2$). Both are supported on $\{O_{(2,2,0,0)},O_{(1,1,1,1)}\}$ and $$R_A=1\cdot\mathbf 1_{O_{(2,2,0,0)}}+(q+1)(q^2+1)\,\mathbf 1_{O_{(1,1,1,1)}},\qquad R_B=(q^2+q)\,\mathbf 1_{O_{(2,2,0,0)}}+0\cdot\mathbf 1_{O_{(1,1,1,1)}} .$$ Proof. Support: every such $M$ satisfies $t\mathcal O^4\subseteq M\subseteq t^{-1}\mathcal O^4$; integrality on $t\mathcal O^4$ gives $\mu_i\ge-2$, and $M=M^*\supseteq(t^{-1}\mathcal O^4)^*=\bigoplus t^{1-\mu_i}\mathcal Oe_i$ with $M\subseteq t^{-1}\mathcal O^4$ gives $\mu_i\le2$; with $\sum\mu_i=4$ only $(2,2,0,0),(1,1,1,1)$ remain. Values: (A, $\mu=(1,1,1,1)$) $tJ$ integral on $M=\mathcal O^4+t^{-1}\tilde P$ iff $\bar J(P,P)=0$: $P$ Lagrangian, $(q+1)(q^2+1)$ of them. (A, $(2,2,0,0)$) pairing $t^{-1}\tilde v$ with $e_3,e_4$ forces $\bar v_3=\bar v_4=0$, so $P=\langle\bar e_1,\bar e_2\rangle$: 1. (B, $(1,1,1,1)$) need $J(L,L)\subseteq t\mathcal O$, but $L\bmod t$ is a hyperplane of $\mathbb F_q^4$, never isotropic: 0. (B, $(2,2,0,0)$) $J_{12}+t^{-2}J_{34}$ integral on $L$ forces $\operatorname{pr}_{34}L=:\Lambda$ with $J(\Lambda,\Lambda)\subseteq t^2\mathcal O$, hence $L=\mathcal O^2\oplus\Lambda$ with $\Lambda\subset\mathcal O^2$ cyclic of index $q^2$: $q^2+q$. Integrality plus the determinant condition gives unimodularity. $\square$ Machine check (script scratchpad/solve/A4/gl4sp4.py, exhaustive enumeration of lattices by Hermite normal form with elementary divisors computed from minors): $q=3$: rows $(1,40)$, $(12,0)$, totals $130=\binom42_3$, $1080$; $q=5$: $(1,156)$, $(30,0)$, totals $806$, $19500$. Agrees with the formulas.

Corollary A4-9.2. $\mathbf 1_{O_{(1,1,1,1)}}=\frac1{(q+1)(q^2+1)}\big(R_A-\tfrac1{q^2+q}R_B\big)\in\mathcal H_K\varphi_0$, for every non-archimedean local $F$ with residue field $\mathbb F_q$ (the lattice argument is characteristic-free). The central elements $T_{m(1,1,1,1)}=\mathbf 1_{t^mK}$ act by $x\mapsto t^{2m}x$, so $\mathbf 1_{O_{(k,k,k,k)}}=T_{\frac{k-1}2(1,1,1,1)}\ast\mathbf 1_{O_{(1,1,1,1)}}\in\mathcal H_K\varphi_0$ for all odd $k$.

Theorem A4-9.3. For $F=\mathbb F_q(\!(t)\!)$, $q$ odd: (SG) holds for $\mathrm{GL}_4/\mathrm{Sp}_4$ (modulo D0, D3). Proof. Prop. B with $f_\mu=T_\lambda$, $\lambda$ a dominant lift, for covered $\mu$ (A4-8's argument), and $f_\mu$ the explicit element of A4-9.2 for uncovered $\mu=(k,k,k,k)$ (support exactly $O_\mu$). $\square$ Hence by Theorem A4-6: $\mathcal C^{\mathrm{Sat}}=\mathcal C$ for $\mathrm{GL}_4/\mathrm{Sp}_4$ over $\overline{\mathbb F}_q$ (D0–D4), and over $\mathbb C$ (D5). In particular $\mathrm{IC}_{(1,1,1,1)}$ is a direct summand of some $V_\lambda\ast\mathrm{IC}_0$ with $\varphi(\lambda)>(1,1,1,1)$ — the retract that the Summary's §4 asked for — and the proof pins the candidates: $\lambda\in\{(1,1,0,0),(1,1,1,-1)\}$ up to $W$ and the convention $\lambda\leftrightarrow\nu$, the Hecke labels of $R_A,R_B$.

Sources

/Users/miltonlin/Projects/Trees/subpages/case-study-1-ramifiedrelative/proof_graphs/spherical-generation/families/symplectic.html (coverage lemma)
/Users/miltonlin/Projects/Trees/subpages/case-study-1-ramifiedrelative/proof_graphs/spherical-generation/classical.html Lemma A(2), Prop. B, Prop. F (orbits as elementary divisors)
scratchpad/solve/A4/gl4sp4.py

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

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 \(\Csat=\cC_K\)

Builds on

Built on by

GL4/Sp4: the uncovered label (1,1,1,1) is a Hecke translate; (SG) over F_q((t)); hence C^Sat = C

Sources