§5 · Agent 5 · Audit: split candidate · 2026-10-17T23:21:24Z

Computation 5.4 · Finite-field test: Hom from Ind of Sp₄(F₁₁) to 1⊠σ_c over SL₂(F₁₁)²

proved

Character sums over Frobenius-fixed flags reproduce the orbit prediction in all 40 cases tested, including the closed-orbit contribution when $\chi_1=1$. This confirms the orbit list, the stabilizers and $c_1=\chi_1/\chi_2$, $c_2=\chi_1\chi_2$.

Computation 5.4.

For $q=11$, $G=\mathrm{Sp}_4(\mathbb F_q)$ on $P\oplus P^\perp$, $H_1=\mathrm{SL}(P)\times\mathrm{SL}(P^\perp)$, $\chi=(\zeta^{a_1},\zeta^{a_2})$ ($\zeta=e^{2\pi i/10}$ on a generator of $\mathbb F_{11}^\times$) and $c=\zeta^k$: $$\dim\operatorname{Hom}_{H_1}(\operatorname{Ind}_B^G\chi,\mathbf 1\boxtimes\sigma_c)=[c\in\{(\chi_1/\chi_2)^{\pm1}\}]+[c\in\{(\chi_1\chi_2)^{\pm1}\}]+[\chi_1=1]\,[\chi_2\in\{c^{\pm1}\}]+[\chi_2=1]\,[\chi_1\in\{c^{\pm1}\}]$$ (each bracket counted with multiplicity 2 when the two values coincide), for $(a_1,a_2)\in\{(1,2),(0,2),(1,3),(2,2)\}$ and all $k\in\mathbb Z/10$.

What was computed.

The character of $\operatorname{Ind}_B^G\chi$. At $h\in H_1$ it is $\sum_{F}\chi_1(\alpha_F)\chi_2(\beta_F)$, summed over flags $F=(\ell\subset L)$ fixed by $h$, where $h$ acts on $\ell$ by $\alpha_F$ and on $L/\ell$ by $\beta_F$. The fixed flags were found from the eigenlines $\ell$ of $h$ and the eigenlines of the induced map on $\ell^\perp/\ell$. As a check, the identity fixes all $17568=(q^4-1)(q+1)/(q-1)$ flags.
Classes. The $q+4=15$ conjugacy classes of $\mathrm{SL}_2(\mathbb F_q)$ are $\pm I$, four unipotent classes, and one class for each trace $\ne\pm2$, split or elliptic. Their sizes sum to $q^3-q$.
Characters of $\sigma_c$. $(q+1)c(\pm1)$ at $\pm I$, $c(\pm1)$ at $\pm$unipotent classes, $c(s)+c(s^{-1})$ at split classes, and $0$ at elliptic classes.
Inner product. $\langle\operatorname{Res}\operatorname{Ind}\chi,\mathbf 1\boxtimes\sigma_c\rangle_{H_1}$, summed over the 225 pairs of classes.

Results. All values are integers to within $10^{-6}$:

$\chi=(\zeta,\zeta^2)$: dimensions $[0,1,0,1,0,0,0,1,0,1]$ for $k=0,\dots,9$. They are nonzero exactly at $c=\zeta^{\pm1}=(\chi_1/\chi_2)^{\pm1}$ and $c=\zeta^{\pm3}=(\chi_1\chi_2)^{\pm1}$.
$\chi=(1,\zeta^2)$: dimension $3$ at $c=\zeta^{\pm2}$, coming from the open, middle and closed orbit $O_a$.
$\chi=(\zeta,\zeta^3)$: $c=\zeta^{\pm2},\zeta^{\pm4}$.
$\chi=(\zeta^2,\zeta^2)$: dimension 2 at $c=1$, since $c_1=1=c_1^{-1}$, and 1 at $c=\zeta^{\pm4}$.

All 40 values match the displayed orbit prediction, which comes from the finite-group Mackey formula $\operatorname{Hom}_{H_1}(\operatorname{Ind}_{S}\mu,\tau)=\operatorname{Hom}_S(\mu,\tau|_S)$ together with $\operatorname{Res}_B\sigma_c=c\oplus\operatorname{Ind}_T^Bc^{-1}$.

Why it suffices (and what it does not test). The test checks the orbit decomposition and the torus characters $(t,t^{-1})$ on $S_0$ and $(t,t)$ on $S_c$, which together determine $c_1,c_2$. It does not test the $p$-adic Ext vanishing or the lifting step of A3-4. Those are formal (see Audit 5.3, steps 3–5).

Builds on

Lemma 3.4 · Generic $I(\chi)$ of $\mathrm{Sp}_4$ has two $\mathrm{SL}_2$-quotients
Audit 5.3 · Audit of A3-4: the Mackey filtration and the two SL₂-quotients of I(χ)

Built on by

Summary 5.12 · Verdict: (SG) fails, proved; $\mathcal C_k^{\mathrm{Sat}}=\mathcal C_k$ modulo (O1)–(O3)

Sources

scratchpad/solve/A5/mackey_ff.py