§4 · Agent 4 · The group: sign and weights · 2026-10-18T18:51:56Z

Proposition 4.10 · Pure sign-type functionals force sheaf failure: obligations for other rows

proved

Abstracts the group argument: any H_I-linear map from C_I to a module killed by e_K whose values on IC traces are traces of pure graded Weil spaces excludes the corresponding simple objects from M_I. This turns each function-level Steinberg or Speh obstruction into one exact sheaf-level obligation, and with (II_I) shows that a divergence can only be ‘functions fail, sheaves hold’.

Proposition 4.10.

Let $X$ be a row of Table 1 over $\overline{\mathbb F}_q$ whose level presentation has $\mathrm{Fr}$-stable pieces with rational points, simple objects $\mathrm{IC}_P$ (constant coefficients) with Weil structures pure of weight $d_P$ ($\mathrm{IC}_P|_P=\overline{\mathbb Q}_\ell[d_P]$), and generators $\mathrm{IC}_w\ast_I\mathrm{IC}_0$ pure and geometrically split as in Setup 4.1(e). Suppose that for every $m\ge1$ there are an $\mathcal H_{I,m}$-module $\rho_m$ with $e_K\rho_m=0$, an $\mathcal H_{I,m}$-linear $\lambda_m:C_c(X(F_m))^I\to\rho_m$, and a linear form $\theta_m$ on $\rho_m$, such that ($\Lambda$) for every simple $\mathrm{IC}_P$: $\theta_m\lambda_m(\operatorname{tr}_m\mathrm{IC}_P)=\sum_j(-1)^j\operatorname{tr}(\mathrm{Fr}^m\mid V_P^j)$ for one finite graded Weil space $V_P^\bullet$ (independent of $m$) with $V_P^j$ pure of weight $d_P+j$. Then $E_{P,n}(w)=0$ for all $w,n$ whenever $V_P^\bullet\ne0$; so every such $\mathrm{IC}_P\notin\mathcal C_I^{\mathrm{Iwa}}$, and if some $V_P^\bullet\ne0$, sheaves fail. (Proved.) For the group, $\rho=$ sign line, $\lambda=\varepsilon$, $\theta=\mathrm{id}$, $V_z=\overline{\mathbb Q}_\ell[0]$ for $z\in\Omega$, $0$ otherwise (Theorem 4.5); the mixed character of Proposition 4.7(2) has $V_z=H^\bullet(\overline{\mathrm{Fl}_z},\mathrm{IC}_z)$.

Generators are killed. $\lambda_m(\varphi_0)=\lambda_m(e_K\ast\varphi_0)=e_K\lambda_m(\varphi_0)=0$; so $\lambda_m(\operatorname{tr}_m(\mathrm{IC}_w\ast_I\mathrm{IC}_0))=\operatorname{tr}_m(\mathrm{IC}_w)\cdot\lambda_m(\varphi_0)=0$ (trace of convolution, as Lemma 4.3(c)). This is Lemma C(1)–(2) of classical.html at the level of traces.
Expansion. As in Lemma 4.4 (its steps 2–5 use only purity, the IC support conditions and a rational point in each piece): $\operatorname{tr}_m(\mathrm{IC}_w\ast\mathrm{IC}_0)=\sum_{P,n}(-1)^n\operatorname{tr}(A_{P,n}^m)\operatorname{tr}_m\mathrm{IC}_P$, eigenvalues of $A_{P,n}$ of weight $W+n-d_P$, $W$ the weight of the generator.
No cancellation. Apply $\theta_m$ and ($\Lambda$): $0=\sum_{P,n,j}(-1)^{n+j}\operatorname{tr}(A^m_{P,n})\operatorname{tr}(\mathrm{Fr}^m|V^j_P)$ for all $m$. An eigenvalue product has weight $(W+n-d_P)+(d_P+j)=W+n+j$, which fixes the sign $(-1)^{n+j}$. Vandermonde as in Theorem 4.5 step 3 gives $\dim E_{P,n}\cdot\dim V_P^j=0$ for all $P,n,j$.
Where $V_P$ can be nonzero. If $\rho_m$ is a sign-type line ($T_s\mapsto-1$) and $\mathrm{IC}_P$ is left $P_s$-equivariant for some simple affine $s$ ($\mathrm{IC}_s\ast\mathrm{IC}_P\simeq\mathrm{IC}_P\otimes H^\bullet(\mathbb P^1)[1]$), the descent argument of Lemma 4.3 step 5 forces $\theta\lambda(\operatorname{tr}\mathrm{IC}_P)=0$. So only pieces that are not $P_s$-saturated for any $s$ can carry $V_P\ne0$: for the group, exactly the closed pieces $I\omega I$ (compare case-group.html, IV.2).
Consequence with (II$_I$). By Corollary 4.6(c) (and its analogue for any $X$ whose inputs D1–D4 hold), (IG) $\Rightarrow\mathcal C_I^{\mathrm{Iwa}}=\mathcal C_I$. So a divergence can only be: (IG) fails and $\mathcal C_I^{\mathrm{Iwa}}=\mathcal C_I$. The proposition says this cannot happen when the function-level obstruction has a lift ($\Lambda$).

Rows (status per row; open unless stated).

$\mathrm{GL}_2/\mathrm{Sp}_2=\mathbb G_m$: $T_s$ acts on $C_I$ by $q$ (det of $I$ is $\mathcal O^\times$), no sign quotient; (IG) holds; sheaves hold (case-gm.html, IV.3). Agree (both hold); not re-proved here.
$\mathrm{GL}_2/T$, $\mathrm{GL}_n/(\mathrm{GL}_p\times\mathrm{GL}_q)$ ($p=q$, $(1,2)$), $\mathrm{Spin}_{2n}/\mathrm{Spin}_{2n-1}$ ($n=2,3$): (IG) fails by a (twisted) Steinberg functional, $\rho$ one-dimensional. Exact obligation: a lift ($\Lambda$) of that functional with some $V_P\ne0$ (then: agree, both fail); without it, divergence remains possible. Heuristic: the candidate pieces are the closed $I$-orbits in $L^+X$ (step 4), matching the pages' expectation that I3 fails there.
$\mathrm{GL}_{2n}/\mathrm{Sp}_{2n}$, $\mathrm{SL}_{2n}/\mathrm{Sp}_{2n}$ ($\operatorname{char}F=0$ Speh obstruction): $\rho$ is not a line; obligation: $\theta$ and a lift ($\Lambda$) of $\theta\lambda$; additionally the Speh argument must be available over $\mathbb F_q(\!(t)\!)$, where (II$_I$) lives (gap: the function failure is stated for $\operatorname{char}F=0$).
$E_6/F_4$: (IG) open; first obligation is the function statement.
$\mathbb A^n$: both hold on the pages; (II$_I$) would give sheaves from (IG) once its inputs D1–D4 are checked at finite levels (not done here; checklist 1 applies).

Builds on

Theorem 4.5 · No $\Delta_\omega$ in any generator $M_{x,y}$; so $\Delta_e\notin\mathcal C_I^{\rm Iwa}$
Corollary 4.6 · (IG) fails by the same character; (II$_I$) holds for the group
Proposition 4.7 · $\mathcal M_I$ for the group: top cells in, $\Omega$ out, exact in rank one

Built on by

Summary 4.11 · Verdict for $X=H$: both levels fail, no divergence

Original record, as the agent wrote it at 18:51 UTC

Pure sign-type functionals force sheaf failure: obligations for other rows

Generators are killed. $\lambda_m(\varphi_0)=\lambda_m(e_K\ast\varphi_0)=e_K\lambda_m(\varphi_0)=0$; so $\lambda_m(\operatorname{tr}_m(\mathrm{IC}_w\ast_I\mathrm{IC}_0))=\operatorname{tr}_m(\mathrm{IC}_w)\cdot\lambda_m(\varphi_0)=0$ (trace of convolution, as Lemma 4.3(c)). This is Lemma C(1)–(2) of classical.html at the level of traces.
Expansion. As in Lemma 4.4 (its steps 2–5 use only purity, the IC support conditions and a rational point in each piece): $\operatorname{tr}_m(\mathrm{IC}_w\ast\mathrm{IC}_0)=\sum_{P,n}(-1)^n\operatorname{tr}(A_{P,n}^m)\operatorname{tr}_m\mathrm{IC}_P$, eigenvalues of $A_{P,n}$ of weight $W+n-d_P$, $W$ the weight of the generator.
No cancellation. Apply $\theta_m$ and ($\Lambda$): $0=\sum_{P,n,j}(-1)^{n+j}\operatorname{tr}(A^m_{P,n})\operatorname{tr}(\mathrm{Fr}^m|V^j_P)$ for all $m$. An eigenvalue product has weight $(W+n-d_P)+(d_P+j)=W+n+j$, which fixes the sign $(-1)^{n+j}$. Vandermonde as in Theorem 4.5 step 3 gives $\dim E_{P,n}\cdot\dim V_P^j=0$ for all $P,n,j$.
Where $V_P$ can be nonzero. If $\rho_m$ is a sign-type line ($T_s\mapsto-1$) and $\mathrm{IC}_P$ is left $P_s$-equivariant for some simple affine $s$ ($\mathrm{IC}_s\ast\mathrm{IC}_P\simeq\mathrm{IC}_P\otimes H^\bullet(\mathbb P^1)[1]$), the descent argument of Lemma 4.3 step 5 forces $\theta\lambda(\operatorname{tr}\mathrm{IC}_P)=0$. So only pieces that are not $P_s$-saturated for any $s$ can carry $V_P\ne0$: for the group, exactly the closed pieces $I\omega I$ (compare case-group.html, IV.2).
Consequence with (II$_I$). By Corollary 4.6(c) (and its analogue for any $X$ whose inputs D1–D4 hold), (IG) $\Rightarrow\mathcal C_I^{\mathrm{Iwa}}=\mathcal C_I$. So a divergence can only be: (IG) fails and $\mathcal C_I^{\mathrm{Iwa}}=\mathcal C_I$. The proposition says this cannot happen when the function-level obstruction has a lift ($\Lambda$).

Rows (status per row; open unless stated).

$\mathrm{GL}_2/\mathrm{Sp}_2=\mathbb G_m$: $T_s$ acts on $C_I$ by $q$ (det of $I$ is $\mathcal O^\times$), no sign quotient; (IG) holds; sheaves hold (case-gm.html, IV.3). Agree (both hold); not re-proved here.
$\mathrm{GL}_2/T$, $\mathrm{GL}_n/(\mathrm{GL}_p\times\mathrm{GL}_q)$ ($p=q$, $(1,2)$), $\mathrm{Spin}_{2n}/\mathrm{Spin}_{2n-1}$ ($n=2,3$): (IG) fails by a (twisted) Steinberg functional, $\rho$ one-dimensional. Exact obligation: a lift ($\Lambda$) of that functional with some $V_P\ne0$ (then: agree, both fail); without it, divergence remains possible. Heuristic: the candidate pieces are the closed $I$-orbits in $L^+X$ (step 4), matching the pages' expectation that I3 fails there.
$\mathrm{GL}_{2n}/\mathrm{Sp}_{2n}$, $\mathrm{SL}_{2n}/\mathrm{Sp}_{2n}$ ($\operatorname{char}F=0$ Speh obstruction): $\rho$ is not a line; obligation: $\theta$ and a lift ($\Lambda$) of $\theta\lambda$; additionally the Speh argument must be available over $\mathbb F_q(\!(t)\!)$, where (II$_I$) lives (gap: the function failure is stated for $\operatorname{char}F=0$).
$E_6/F_4$: (IG) open; first obligation is the function statement.
$\mathbb A^n$: both hold on the pages; (II$_I$) would give sheaves from (IG) once its inputs D1–D4 are checked at finite levels (not done here; checklist 1 applies).

Sources

classical.html, Lemma C, Corollary D, Propositions E, F
index.html, Table 1
case-gm.html, Proposition IV.3
case-group.html, Proposition IV.2

Table 1. Cumulative token usage and estimated cost.
Work	Tokens	US $
Original search, audits and page construction	384.00 M	≈ $171.00
Presentation edits and follow-up corrections	9.64 M	≈ $4.99
Presentation edits and follow-up corrections	9.82 M	≈ $4.29
Total	≈ 403.46 M	≈ $180.28