§4 · Agent 4 · Audit: counterexample and mechanism · 2026-10-17T22:12:40Z

Computation 4.6 · Computation: hermitian Hecke counts $T_\lambda\ast\varphi_0$ for $q=3,5$

proved

All explicit identities of Computation 2.7 were recomputed exactly for $q=3$ and $q=5$ by an independent coset enumeration, and extended to $T_{(2,1)}$, $T_{(3,0)}$. A separate volume identity confirms the coefficient $q+1$. The codimension of the Hecke span on $L_d$ is $1,1,2$ for $d=1,2,3$, as Theorem 2.6(5) predicts.

Computation 4.6.

For $q=3$ and $q=5$, with $\mathbf 1_c$ the indicator of the $K$-orbit of type $c$: $$T_{(1,0)}\ast\varphi_0=\mathbf 1_{(2,0)}+(q+1)\mathbf 1_{(1,1)},\qquad T_{(1,1)}\ast\varphi_0=\mathbf 1_{(2,2)},\qquad T_{(2,1)}\ast\varphi_0=\mathbf 1_{(4,2)}+(q+1)\mathbf 1_{(3,3)},$$ $$T_{(2,0)}\ast\varphi_0=\mathbf 1_{(4,0)}+(q+1)\mathbf 1_{(3,1)}+q(q+1)\mathbf 1_{(2,2)},\qquad T_{(1,-1)}\ast\varphi_0=\mathbf 1_{(2,-2)}+(q+1)\mathbf 1_{(1,-1)}+q(q+1)\mathbf 1_{(0,0)},$$ and for $q=3$, $T_{(3,0)}\ast\varphi_0=\mathbf 1_{(6,0)}+4\,\mathbf 1_{(5,1)}+12\,\mathbf 1_{(4,2)}+36\,\mathbf 1_{(3,3)}$. Every value vanishes on odd $c_1+c_2$ and outside $\{c\le2\lambda\}$. For $\mathrm{SL}_2$: $(T_{\alpha^\vee}\ast\varphi_0)(\operatorname{diag}(t^n,t^{-n}))=q(q+1),\,q+1,\,1,\,0$ for $n=0,1,2,3$. On $L_d=\{c\le(2d,0)\}$ the Hecke span has codimension $1,1,2$ for $d=1,2,3$ ($q=3$), equal to $p_{\le2}(2d)-p_{\le2}(d)$.

Outcome: Computation 2.7 holds exactly; Theorem 2.6(5) confirmed for $n=2$, $d\le3$.

Formula. $(T_\lambda\ast\varphi_0)(x)=\#\{gK\subseteq Kt^\lambda K:\ g^{-1}xg^{-*}\in\mathrm{GL}_2(\mathcal O_E)\}$. Coset representatives (Hermite normal form): $g=t^{\lambda_2}\left(\begin{smallmatrix}t^i&b\\0&t^j\end{smallmatrix}\right)$, $i+j=m:=\lambda_1-\lambda_2$, $b\in\mathbb F_{q^2}[t]$ of degree $<i$, $\min(i,j,v(b))=0$. There are $q^{2m}+q^{2m-2}$ of them for $m\ge1$ ($10,90,810$ for $q=3$; $26,650$ for $q=5$). The test was done in exact arithmetic over $\mathbb F_{q^2}[t,t^{-1}]$, $\mathbb F_{q^2}=\mathbb F_q(\sqrt2)$, at $x=\operatorname{diag}(t^{c_1},t^{c_2})$ for every dominant $(c_1,c_2)\in[-3,6]^2$; by Audit 4.3 these represent all $K$-orbits in that range.
Two entries by hand. At $x=tI$ and $g=\left(\begin{smallmatrix}t&b\\0&1\end{smallmatrix}\right)$, $b\in\mathbb F_{q^2}$: $g^{-1}xg^{-*}=\left(\begin{smallmatrix}t^{-1}(1+N b)&-b\\-\bar b&t\end{smallmatrix}\right)$, integral iff $Nb=-1$ ($q+1$ solutions), and then it equals $\left(\begin{smallmatrix}0&-b\\-\bar b&t\end{smallmatrix}\right)$, of determinant $-Nb=1$, so unimodular; the coset of $\operatorname{diag}(1,t)$ gives $\operatorname{diag}(t,t^{-1})$, not integral. At $x=\operatorname{diag}(t^2,1)$ only $b=0$ works.
Independent volume check of the first identity. Let $\mu$ be a $G(F)$-invariant measure on the base $G(F)$-orbit and $U=U(1_2)$. Integrating gives $\int T_{(1,0)}\ast\varphi_0\,d\mu=\#(Kt^{(1,0)}K/K)\,\mu(O_0)=(q^2+1)\mu(O_0)$. For $\operatorname{diag}(t^2,1)=gg^*$, $g=\operatorname{diag}(t,1)$: $g^{-1}Kg\cap U$ is the stabilizer of $M=t^{-1}\mathcal O_Ee_1\oplus\mathcal O_Ee_2$; the only unimodular lattice between $M^\vee=t\mathcal O_Ee_1\oplus\mathcal O_Ee_2$ and $M$ is $\mathcal O_E^2$, so this stabilizer is $\{u\in U(\mathcal O):u_{21}\in t\mathcal O_E\}$, of index $q^2-q$ (the anisotropic lines of the hermitian plane over $\mathbb F_{q^2}$): $\mu(O_{(2,0)})=(q^2-q)\mu(O_0)$. For $tI=g'g'^*$, $g'=\left(\begin{smallmatrix}a&1\\-1&\bar a\end{smallmatrix}\right)$, $Na=t-1$: $g'$ is a similitude normalizing $U$, so $\mu(O_{(1,1)})=\mu(O_0)$. Then $\mu(O_{(2,0)})+(q+1)\mu(O_{(1,1)})=(q^2-q)+(q+1)=q^2+1$: consistent.
Ranks. On $L_d$ (coordinates ordered $(2d,0),(2d-1,1),\dots$) the vectors $T_a\ast\varphi_0$, $2a\in L_d$, are $(1,4)$ for $d=1$; $(1,4,12),(0,0,1)$ for $d=2$; $(1,4,12,36),(0,0,1,4)$ for $d=3$ ($q=3$); similarly with $6,30$ for $q=5$. Codimensions $1,1,2=p_{\le2}(2d)-p_{\le2}(d)$; $\mathbf 1_{(2d-1,1)}$ is never in the span.
$\mathrm{SL}_2$. The representatives $t^{-1}\left(\begin{smallmatrix}t^i&b\\0&t^j\end{smallmatrix}\right)$, $i+j=2$, have determinant $1$ and represent the $K_{\mathrm{SL}}$-cosets of $K_{\mathrm{SL}}t^{\alpha^\vee}K_{\mathrm{SL}}$ (the natural map to $Kt^{(1,-1)}K/K$ is injective since $K\cap\mathrm{SL}_2(E)=K_{\mathrm{SL}}$, and surjective by adjusting with $\operatorname{diag}(u,1)$, which commutes with $t^{(1,-1)}$). Values at $\operatorname{diag}(t^n,t^{-n})$, $n=0,1,2,3$: $12,4,1,0$ ($q=3$) and $30,6,1,0$ ($q=5$).
Pattern (observation). $T_{(m,0)}\ast\varphi_0=\mathbf 1_{(2m,0)}+\sum_{k=1}^m(q+1)q^{k-1}\mathbf 1_{(2m-k,k)}$ for $m\le3$, $q=3$: sphere sizes of the $(q+1)$-regular tree of $U(1,1)(F)$, on which the unimodular lattices form one vertex type.

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Computation 2.7 · Computation: $T_\lambda\ast\varphi_0$ on hermitian $2\times2$ matrices
Theorem 2.6 · Theorem: the unitary twist of $(H\times H)/H$ fails (SG) over $\mathbb F_q(\!(t)\!)$

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Summary 4.12 · Verdict of the audit: the unitary twist is a valid counterexample; the repairs are minor

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script herm_audit.py (q = 3, 5)

Computation 4.6 · Computation: hermitian Hecke counts \(T_\lambda\ast\varphi_0\) for \(q=3,5\)

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