§5 · Agent 5 · 𝔸ⁿ: lattices · 2026-10-18T18:42:39Z

Proposition 5.6 · (IG) for $\mathbb A^n$: $T_{\tau^k}\ast\varphi_0=\mathbf 1_{\tau^k\mathcal O^n}$

proved ★ key step

With $\operatorname{vol}I=1$, the length-zero Hecke operators carry $\varphi_0$ to the lattice indicators $b_k=\mathbf 1_{\tau^k\mathcal O^n}$, which form a basis of $C_I$; so (IG) holds over every non-archimedean $F$, already for the commutative subalgebra $\mathbb C[\Omega]$, and the full $\mathcal H_I$-action is explicit. Over $\mathbb F_q((t))$ this is the trace of Computation 5.3.

Proposition 5.6.

Let $F$ be non-archimedean with $\mathcal O$, uniformizer $t$, residue field $\mathbb F_q$; $\tau=t^{\varepsilon_n}\sigma\in G(F)$; $\operatorname{vol}(I)=1$, $T_w=\mathbf 1_{IwI}$, quadratic relation $(T_s-q)(T_s+1)=0$. Put $b_k:=\mathbf 1_{\tau^k\mathcal O^n}\in C_I$. Then

(a) $T_{\tau^k}\ast\varphi_0=b_k$ and $T_\tau\ast b_k=b_{k+1}$;

(b) for $n\ge2$: $T_{s_i}\ast b_k=q\,b_k$ if $i\not\equiv-k\pmod n$, and $T_{s_i}\ast b_k=b_{k-1}+q\,b_{k+1}-b_k$ if $i\equiv-k$;

(c) $\{b_k\}{k\in\mathbb Z}$ is a basis of $C_I$, so $h\mapsto h\ast\varphi_0$ is an isomorphism $\mathbb C[\Omega]=\bigoplus_k\mathbb C\,T{\tau^k}\xrightarrow{\sim}C_I$ and $\boxed{\mathcal H_I\ast\varphi_0=\mathbb C[\Omega]\ast\varphi_0=C_I}$;

(d) $\mathbf 1_{O_{m,j}}=(T_{\tau^k}-T_{\tau^{k+1}})\ast\varphi_0$ with $k=nm+n-j$; and, comparing normalizations, $T_{t^{\lambda_{m,j}}}\ast\varphi_0=q^{j(n-j)}b_k$ (IV.1) and $\mathbf 1_{g_{m,j}K}\ast\varphi_0=[K:I]\,b_k$ (F3, Lemma 2);

(e) for $F=\mathbb F_q((t))$ and canonical Weil structures: $\operatorname{tr}(\mathrm{IC}{\tau^k}\ast\mathrm{IC}_0)=b_k$, $\operatorname{tr}(S_k)=(-1)^kb_k$, and $\operatorname{tr}(\mathrm{IC}{s_0}\ast\mathrm{IC}0)=-(1+T{s_0})\ast\varphi_0=-b_{-1}-q\,b_1$, the traces of Computation 5.3(a),(c).

Proof.

  1. (a) $I\tau^kI=\tau^kI$ (Lemma 5.1(b)) and $\operatorname{vol}I=1$, so $(T_{\tau^k}\ast f)(x)=\int_{\tau^kI}f(g^{-1}x)\,dg=f(\tau^{-k}x)$, i.e. $T_{\tau^k}\ast\mathbf 1_L=\mathbf 1_{\tau^kL}$; take $L=\mathcal O^n$, resp. $L=\tau^k\mathcal O^n$ and $k=1$.
  1. (b) $\tau^k\mathcal O^n=\Lambda_{m,j}$ with $j\equiv-k\pmod n$ ($j=n$ read as $0$). If $i\not\equiv j$, $P_{s_i}$ fixes $\Lambda_{m,j}$, and for a $P_s$-stable lattice $T_s\ast\mathbf 1_L=|IsI/I|\,\mathbf 1_L=q\,\mathbf 1_L$. If $i\equiv-k$: since $\ell(\tau)=0$, $T_{s_i}T_{\tau^k}=T_{s_i\tau^k}=T_{\tau^k}T_{s_{i+k}}=T_{\tau^k}T_{s_0}$, and $T_{s_0}\ast\varphi_0=\sum_{h\in Is_0I/I}\mathbf 1_{h\mathcal O^n}$ is the sum of $\mathbf 1_L$ over the $q$ lattices $L\ne\mathcal O^n$ of the $\mathbb P^1(\mathbb F_q)$ in Computation 5.3, step 3. A vector of $\tau^{-1}\mathcal O^n\setminus\tau\mathcal O^n$ lies in exactly one of the $q+1$ lattices, a vector of $\tau\mathcal O^n$ in all of them, so $\sum_{\text{all }L}\mathbf 1_L=b_{-1}+q\,b_1$ and $T_{s_0}\ast\varphi_0=b_{-1}+q\,b_1-b_0$; apply $T_{\tau^k}$. Check of the quadratic relation: $b_{\pm1}$ are $P_{s_0}$-stable, so $T_{s_0}(b_{-1}+qb_1-b_0)=(q-1)(b_{-1}+qb_1-b_0)+q\,b_0$, i.e. $T_{s_0}^2=(q-1)T_{s_0}+q$ on $\varphi_0$. Computation: for $n=2,3$, $q=3$, the value of $T_{s_0}\ast\varphi_0$ was computed at all $3^{3n}$ points of $t^{-1}\mathcal O^n/t^2\mathcal O^n$ from the coset representatives $(1+ctE_{n1})s_0$, $c\in\mathbb F_q$, of $Is_0I/I$; it agrees with $b_{-1}+qb_1-b_0$ everywhere.
  1. (c) Independence: the $\tau^k\mathcal O^n$ are strictly decreasing; evaluate on $\tau^k\mathcal O^n\setminus\tau^{k+1}\mathcal O^n$. Spanning: $\varphi\in C_I$ is supported on some $t^{-N}\mathcal O^n=\tau^{-nN}\mathcal O^n$, constant on some $t^M\mathcal O^n=\tau^{nM}\mathcal O^n$ (local constancy at $0$), and constant on each orbit $o_k=\tau^k\mathcal O^n\setminus\tau^{k+1}\mathcal O^n$ (Lemma 5.1(c)); so $\varphi=\sum_{k=-nN}^{nM-1}\varphi(o_k)(b_k-b_{k+1})+\varphi(0)\,b_{nM}$. With (a), $T_{\tau^k}\mapsto b_k$ is a bijection of bases.
  1. (d) $O_{m,j}=o_{k(m,j)}$ (Lemma 5.1(c)) and (a). $It^\lambda I\subseteq t^\lambda K$ consists of $q^{\ell(t^\lambda)}$ cosets $hI$, each with $h\mathcal O^n=t^\lambda\mathcal O^n$, and $\ell(t^{\lambda_{m,j}})=j(n-j)$ (Computation 5.3(b)); $g_{m,j}K$ consists of $[K:I]$ such cosets.
  1. (e) $\delta_{\tau^k}$ and $\underline{\overline{\mathbb Q}}\ell$ on $\tau^k\mathcal O^n$ (defined over $\mathbb F_q$) have trivial Frobenius on stalks, so the traces are $T{\tau^k}$ and $b_k$; $S_k$, the constant sheaf on $\tau^k\mathcal O^n$ shifted by $-k$, has trace $(-1)^kb_k$. For $s_0$: $\mathrm{IC}{s_0}=\overline{\mathbb Q}\ell[1]$ on $\mathbb P^1$ has trace $-(1+T_{s_0})$, and $m_!$ in Computation 5.3, step 3, is $\overline{\mathbb Q}{\ell,\tau^{-1}\mathcal O^n}[1]\oplus\overline{\mathbb Q}{\ell,\tau\mathcal O^n}[-1](-1)$, with trace $-b_{-1}-q\,b_1=-(1+T_{s_0})\ast\varphi_0$ by step 2. For the generators $\delta_{\tau^k}$, trace of convolution = convolution of traces is just the Frobenius-equivariance of $x\mapsto\tau^kx$; no Grothendieck–Lefschetz argument is needed for (IG) by traces.

Comparison with F3. F3 proves (IG) through $[K:I]^{-1}(\mathbf 1_{g_{m,j}K}-\mathbf 1_{g_{m,j-1}K})\ast\varphi_0$. New here: no normalizing constants; the cyclic generation over the commutative subalgebra $\mathbb C[\Omega]\simeq\mathbb C[z^{\pm1}]$, with $C_I$ free of rank one; and the explicit $T_{s_i}$-action (b), which is the function shadow of Computation 5.3(c).

Builds on

  • Lemma 5.1 · Orbits of $I$ on $L\mathbb A^n$: closures $\tau^k\mathcal O^n$, $\tau\in\Omega$
  • Computation 5.3 · $\mathrm{IC}_{\tau^k}\ast_I\mathrm{IC}_0=S_k[k]$ and the simple reflections

Built on by

  • Corollary 5.8 · Reverse route for $\mathbb A^n$: (IG) and (II$_I$) give sheaf generation
  • Summary 5.9 · Verdict: for $\mathbb A^n$ both Iwahori-level statements hold
  • Audit 8.10 · Audit of (IG) for $\mathbb A^n$ and the reverse route (A5-6, A5-8)
  • Strategy 10.1 · Registry after round one: every target has a candidate proof; three adversarial audits are launched
  • Summary 10.2 · Verdict: the criterion and the bridge hold; the decided rows of Table 1 agree; GL₂/T fails at both levels
Original record, as the agent wrote it at 18:42 UTC

(IG) for $\mathbb A^n$: $T_{\tau^k}\ast\varphi_0=\mathbf 1_{\tau^k\mathcal O^n}$

Proof.

  1. (a) $I\tau^kI=\tau^kI$ (Lemma 5.1(b)) and $\operatorname{vol}I=1$, so $(T_{\tau^k}\ast f)(x)=\int_{\tau^kI}f(g^{-1}x)\,dg=f(\tau^{-k}x)$, i.e. $T_{\tau^k}\ast\mathbf 1_L=\mathbf 1_{\tau^kL}$; take $L=\mathcal O^n$, resp. $L=\tau^k\mathcal O^n$ and $k=1$.
  1. (b) $\tau^k\mathcal O^n=\Lambda_{m,j}$ with $j\equiv-k\pmod n$ ($j=n$ read as $0$). If $i\not\equiv j$, $P_{s_i}$ fixes $\Lambda_{m,j}$, and for a $P_s$-stable lattice $T_s\ast\mathbf 1_L=|IsI/I|\,\mathbf 1_L=q\,\mathbf 1_L$. If $i\equiv-k$: since $\ell(\tau)=0$, $T_{s_i}T_{\tau^k}=T_{s_i\tau^k}=T_{\tau^k}T_{s_{i+k}}=T_{\tau^k}T_{s_0}$, and $T_{s_0}\ast\varphi_0=\sum_{h\in Is_0I/I}\mathbf 1_{h\mathcal O^n}$ is the sum of $\mathbf 1_L$ over the $q$ lattices $L\ne\mathcal O^n$ of the $\mathbb P^1(\mathbb F_q)$ in Computation 5.3, step 3. A vector of $\tau^{-1}\mathcal O^n\setminus\tau\mathcal O^n$ lies in exactly one of the $q+1$ lattices, a vector of $\tau\mathcal O^n$ in all of them, so $\sum_{\text{all }L}\mathbf 1_L=b_{-1}+q\,b_1$ and $T_{s_0}\ast\varphi_0=b_{-1}+q\,b_1-b_0$; apply $T_{\tau^k}$. Check of the quadratic relation: $b_{\pm1}$ are $P_{s_0}$-stable, so $T_{s_0}(b_{-1}+qb_1-b_0)=(q-1)(b_{-1}+qb_1-b_0)+q\,b_0$, i.e. $T_{s_0}^2=(q-1)T_{s_0}+q$ on $\varphi_0$. Computation: for $n=2,3$, $q=3$, the value of $T_{s_0}\ast\varphi_0$ was computed at all $3^{3n}$ points of $t^{-1}\mathcal O^n/t^2\mathcal O^n$ from the coset representatives $(1+ctE_{n1})s_0$, $c\in\mathbb F_q$, of $Is_0I/I$; it agrees with $b_{-1}+qb_1-b_0$ everywhere.
  1. (c) Independence: the $\tau^k\mathcal O^n$ are strictly decreasing; evaluate on $\tau^k\mathcal O^n\setminus\tau^{k+1}\mathcal O^n$. Spanning: $\varphi\in C_I$ is supported on some $t^{-N}\mathcal O^n=\tau^{-nN}\mathcal O^n$, constant on some $t^M\mathcal O^n=\tau^{nM}\mathcal O^n$ (local constancy at $0$), and constant on each orbit $o_k=\tau^k\mathcal O^n\setminus\tau^{k+1}\mathcal O^n$ (Lemma 5.1(c)); so $\varphi=\sum_{k=-nN}^{nM-1}\varphi(o_k)(b_k-b_{k+1})+\varphi(0)\,b_{nM}$. With (a), $T_{\tau^k}\mapsto b_k$ is a bijection of bases.
  1. (d) $O_{m,j}=o_{k(m,j)}$ (Lemma 5.1(c)) and (a). $It^\lambda I\subseteq t^\lambda K$ consists of $q^{\ell(t^\lambda)}$ cosets $hI$, each with $h\mathcal O^n=t^\lambda\mathcal O^n$, and $\ell(t^{\lambda_{m,j}})=j(n-j)$ (Computation 5.3(b)); $g_{m,j}K$ consists of $[K:I]$ such cosets.
  1. (e) $\delta_{\tau^k}$ and $\underline{\overline{\mathbb Q}}\ell$ on $\tau^k\mathcal O^n$ (defined over $\mathbb F_q$) have trivial Frobenius on stalks, so the traces are $T{\tau^k}$ and $b_k$; $S_k$, the constant sheaf on $\tau^k\mathcal O^n$ shifted by $-k$, has trace $(-1)^kb_k$. For $s_0$: $\mathrm{IC}{s_0}=\overline{\mathbb Q}\ell[1]$ on $\mathbb P^1$ has trace $-(1+T_{s_0})$, and $m_!$ in Computation 5.3, step 3, is $\overline{\mathbb Q}{\ell,\tau^{-1}\mathcal O^n}[1]\oplus\overline{\mathbb Q}{\ell,\tau\mathcal O^n}[-1](-1)$, with trace $-b_{-1}-q\,b_1=-(1+T_{s_0})\ast\varphi_0$ by step 2. For the generators $\delta_{\tau^k}$, trace of convolution = convolution of traces is just the Frobenius-equivariance of $x\mapsto\tau^kx$; no Grothendieck–Lefschetz argument is needed for (IG) by traces.

Comparison with F3. F3 proves (IG) through $[K:I]^{-1}(\mathbf 1_{g_{m,j}K}-\mathbf 1_{g_{m,j-1}K})\ast\varphi_0$. New here: no normalizing constants; the cyclic generation over the commutative subalgebra $\mathbb C[\Omega]\simeq\mathbb C[z^{\pm1}]$, with $C_I$ free of rank one; and the explicit $T_{s_i}$-action (b), which is the function shadow of Computation 5.3(c).

Sources