§5 · Agent 5 · Functions without (A1)–(A5) · 2026-10-17T03:32:17Z

Lemma 5.6 · \(\GL_{2m}/\Sp_{2m}\): the rows \(\lambda^*(b)=(b,0)\) are triangular for reverse dominance

proved audit: error

Error. Lemmas 1–3 are correct: re-derived step by step, and confirmed exhaustively by counting Lagrangian submodules of \(\Lambda_0^\#/\Lambda_0\) (every label with \(|b|\le5\), \(m\le4\) over \(\F_2[[t]]\); \(|b|\le4\), \(m\le3\) over \(\F_3[[t]]\)). The \(m=3\) check table, stated as proved hand counts, contains two false entries and leaves one undetermined: \(N(\lambda^*(3,1,0),(3,1,0))=q^3(q+1)^2\), not \(q^2(q+1)^2\) (\(72\) versus \(36\) at \(q=2\), \(432\) versus \(144\) at \(q=3\)); \(N(\lambda^*(2,1,1),(2,1,1))=q^3(q+1)^2(q^2+1)\), not \(q(q+1)^2(q^2+1)\) (\(360\) versus \(90\), \(4320\) versus \(480\)); the entry \(\ast\) is \((q+1)^2\). The hand counts kept only the Lagrangians that split along the orthogonal decomposition of \(\Lambda_0^\#/\Lambda_0\) and missed those that are graphs of anti-isometries. All other entries, and the zero pattern, are right, so the agreement with Lemma 3 stands. Corrected in Correction 10.6. The conclusion still holds by another argument.
Corrected by: Correction 10.6 · Correction: the \(m=3\) check table for \(\GL_6/\Sp_6\), with the Lagrangians that are graphs, Correction X1.2 · GL₆/Sp₆: Lemmas 1–3 re-proved; the |b|=4 table proved for all q; |b|≤5 computed.

For the Hecke row \(\lambda^*(b)=(b_1,\dots,b_m,0,\dots,0)\) the translate of the basic function is nonzero at the orbit \(b\) and vanishes at every orbit that is not above \(b\) in dominance; the vanishing comes from a submodule-type inequality for the self-dual lattice. This triangularity replaces the missing dominant lifts. The node also records a hand-computed check table for \(m=3\).

Lemma 5.6.

Notation of Computation 5.5 with \(m\) blocks: \(b=(b_1\ge\dots\ge b_m)\), \(x_b=\bigoplus_i\varpi^{b_i}J_2\), \(N(\lambda,b)=(T_\lambda\ast\varphi_0)(x_b)\); \(F\) is any non-archimedean local field.

  1. (Lemma 1) \(T_{(d,0,\dots,0)}\ast\varphi_0=q^{d-1}(q+1)\,\one_{O_{(d,0,\dots,0)}}\) for \(d\ge1\).
  2. (Lemma 2) For \(b\ge0\) and \(\lambda^*(b):=(b_1,\dots,b_m,0,\dots,0)\in\Z^{2m}\): \(N(\lambda^*(b),b)\ge1\).
  3. (Lemma 3) If \(N(\lambda^*(b),b')\ne0\), then \(b'\ge0\), \(|b'|=|b|\) and \(b'\ge_{\mathrm{dom}}b\) (every partial sum of \(b'\) is at least the corresponding partial sum of \(b\)).
  4. (Check, \(m=3\), \(|b|=4\)) Hand counts, columns \(b'=(4,0,0),(3,1,0),(2,2,0),(2,1,1)\):
Table 2. row \(\lambda\) and \((4,0,0)\) and \((3,1,0)\)
row \(\lambda\)\((4,0,0)\)\((3,1,0)\)\((2,2,0)\)\((2,1,1)\)
\((4,0,0,0,0,0)\)\(q^3(q+1)\)\(0\)\(0\)\(0\)
\((3,1,0,0,0,0)\)\(q(q+1)\)\(q^2(q+1)^2\)\(0\)\(0\)
\((2,2,0,0,0,0)\)\(1\)\(q(q^2-1)\)\(q^3(q+1)(q^2+1)\)\(0\)
\((2,1,1,0,0,0)\)\(0\)\(\ast\)\(q(q+1)(q^2+1)\)\(q(q+1)^2(q^2+1)\)
\((1,1,1,1,0,0)\)\(0\)\(0\)\(1\)\((q+1)(q^2+1)\)

Every nonzero entry in a row \(\lambda^*(b)\) lies in a column \(b'\ge_{\mathrm{dom}}b\), as Lemma 3 predicts.

Proof of Lemma 1. Here \(\Lambda''=\Lambda_0+\varpi^{-d}\cO\tilde u\) with \(\tilde u\) primitive. Integrality of \(x_b\) on \(\Lambda''\) means \(x_b(\tilde u,\Lambda_0)\subseteq\varpi^d\cO\), that is, the \(i\)-th block of \(\tilde u\) lies in \(\varpi^{\max(d-b_i,0)}\cO^2\). Primitivity forces \(b_1\ge d\); together with \(|b|=d\) and \(b\ge0\) this gives \(b=(d,0,\dots,0)\), and then the admissible lattices correspond to the points of \(\mathbb P^1(\cO/\varpi^d)\), of which there are \(q^{d-1}(q+1)\).

Proof of Lemma 2. Take \(\Lambda''=\bigoplus_i(\varpi^{-b_i}\cO e_{2i-1}\oplus\cO e_{2i})\supseteq\Lambda_0\). Then \(\Lambda''/\Lambda_0\cong\bigoplus_i\cO/\varpi^{b_i}\) has type \(\lambda^*(b)\), and \(x_b(\varpi^{-b_i}e_{2i-1},e_{2i})=1\), so \(x_b\) is unimodular on \(\Lambda''\).

Proof of Lemma 3. Let \(\Lambda_0\subseteq\Lambda''\) with \(M:=\Lambda''/\Lambda_0\) of type \(b\), let \(x\) be unimodular on \(\Lambda''\), and suppose \(x|_{\Lambda_0}\cong x_{b'}\).

  1. \(b'\ge0\) by integrality on \(\Lambda_0\), and \(|b'|=|b|\) by the Pfaffian.
  2. Put \(\Lambda_0^\#=\{v:x(v,\Lambda_0)\subseteq\cO\}\). Since \(\Lambda_0\subseteq\Lambda''=\Lambda''^\#\), we have \(\Lambda''\subseteq\Lambda_0^\#\). In a symplectic basis of \(\Lambda_0\), \(M':=\Lambda_0^\#/\Lambda_0\cong\bigoplus_i(\cO/\varpi^{b'_i})^2\), of type \(b'^{(2)}=(b'_1,b'_1,\dots,b'_m,b'_m)\).
  3. \(x\) induces a perfect pairing \(\Lambda_0^\#/\Lambda''\times\Lambda''/\Lambda_0\to F/\cO\), so \(N:=\Lambda_0^\#/\Lambda''\cong\Hom(M,F/\cO)\) also has type \(b\), and \(0\to M\to M'\to N\to0\) is exact.
  4. For a finite \(\cO\)-module of type \(t\), \(\sum_{i\le k}t_i\) is the largest value of \(\log_q|S|\) over submodules \(S\) generated by \(k\) elements. Take \(S_1\subseteq M\) and \(S_2\subseteq N\) optimal with \(k\) generators each, and let \(S\) be \(S_1\) plus lifts of generators of \(S_2\): a submodule of \(M'\) with \(2k\) generators and \(|S|\ge|S_1||S_2|\). Hence
$$2\sum_{i\le k}b'_i=\sum_{i\le2k}b'^{(2)}_i\ \ge\ 2\sum_{i\le k}b_i ,$$

and \(b'\ge_{\mathrm{dom}}b\). \(\square\)

The check table. The \(m=3\) entries were obtained by the same block-by-block integrality analysis, by hand and without machine check. The \(m=2\) data of Computation 5.5 also obey the pattern of Lemma 3.

Builds on

  • Computation 5.5 · \(\GL_4/\Sp_4\): lattice counts and a Hecke element for the uncovered label \((1,1)\)

Built on by

  • Theorem 5.7 · \((\mathrm{SG})\) for \(\GL_{2m}/\Sp_{2m}\) (all \(m\)) and \(\SL_{2n}/\Sp_{2n}\) (all \(n\)), every non-archimedean local field
  • Audit 7.22 · Audit of the \(\GL_4/\Sp_4\) counts and of Lemmas 1–3; two rows of the \(m=3\) table recomputed over \(\F_2[[t]]\) and \(\F_3[[t]]\)
  • Correction 10.6 · Correction: the \(m=3\) check table for \(\GL_6/\Sp_6\), with the Lagrangians that are graphs
  • Correction X1.2 · GL₆/Sp₆: Lemmas 1–3 re-proved; the |b|=4 table proved for all q; |b|≤5 computed
Original record, as the agent wrote it at 03:32 UTC

GL_{2m}/Sp_{2m}: the row lambda*(b)=(b,0) is triangular for reverse dominance

Notation of node 5 with \(m\) blocks; \(b=(b_1\ge\dots\ge b_m)\), \(x_b=\oplus\varpi^{b_i}J_2\), \(N(\lambda,b)\) as before (any \(F\)).

Lemma 1 (central/diagonal rows). \(T_{(d,0,\dots,0)}\ast\varphi_0=q^{d-1}(q+1)\,\mathbf 1_{O_{(d,0,\dots,0)}}\): for \(\Lambda''=\Lambda_0+\varpi^{-d}\mathcal O\tilde u\), integrality gives \(\tilde u^{(i)}\in\varpi^{d-b_i}\mathcal O^2\) per block, primitivity forces \(b_1\ge d\), so \(b=(d,0,\dots)\), and the count is \(|\mathbb P^1(\mathcal O/\varpi^d)|\).

Lemma 2 (nonvanishing). For \(\lambda^*(b):=(b_1,\dots,b_m,0,\dots,0)\) (\(b\ge0\)): \(N(\lambda^*(b),b)\ge1\). Transposed count: in the standard unimodular \((\Lambda'',x)\), \(\Lambda_0:=\oplus_i(\varpi^{b_i}\mathcal Oe_{2i-1}\oplus\mathcal Oe_{2i})\) has \(\Lambda''/\Lambda_0\cong\oplus\mathcal O/\varpi^{b_i}\) and \(x|_{\Lambda_0}\cong x_b\).

Lemma 3 (vanishing, (V)). \(N(\lambda^*(b),b')\ne0\ \Rightarrow\ b'\ge_{\mathrm{dom}}b\) (partial sums), with \(b'\ge0\), \(|b'|=|b|\). Proof. Given \(\Lambda_0\subseteq\Lambda''\) with \(M:=\Lambda''/\Lambda_0\) of type \(b\) and \(x|_{\Lambda_0}\cong x_{b'}\), let \(\Lambda_0^\#=\{v:x(v,\Lambda_0)\subseteq\mathcal O\}\supseteq\Lambda''=\Lambda''^\#\). Then \(M':=\Lambda_0^\#/\Lambda_0\cong\oplus_i(\mathcal O/\varpi^{b'_i})^2\) (symplectic basis of \(\Lambda_0\)), type \(b'^{(2)}:=(b'_1,b'_1,\dots,b'_m,b'_m)\), and \(0\to M\to M'\to N\to0\) with \(N=\Lambda_0^\#/\Lambda''\cong\mathrm{Hom}(M,F/\mathcal O)\), also of type \(b\). For finite \(\mathcal O\)-modules, \(\sum_{i\le k}\mathrm{type}(M')_i=\max\{\log_q|S|:S\subseteq M'\text{ generated by }k\text{ elements}\}\); taking \(S=S_1+\)(lifts of generators of \(S_2\)) with \(S_1\subseteq M\), \(S_2\subseteq N\) gives \(\mathrm{type}(M')\ge_{\mathrm{dom}}\mathrm{type}(M)\cup\mathrm{type}(N)=b\cup b=b^{(2)}\). Even partial sums: \(2\sum_{i\le k}b'_i\ge2\sum_{i\le k}b_i\). \(\square\)

Checks against hand counts (\(m=3\), \(|b|=4\), columns \((4,0,0),(3,1,0),(2,2,0),(2,1,1)\)): rows \(\lambda^*\!=(4,0^5)\): \((q^3(q+1),0,0,0)\); \((3,1,0^4)\): \((q(q+1),q^2(q+1)^2,0,0)\); \((2,2,0^4)\): \((1,q(q^2-1),q^3(q+1)(q^2+1),0)\); \((2,1,1,0^3)\): \((0,\ast,q(q+1)(q^2+1),q(q+1)^2(q^2+1))\); \((1,1,1,1,0,0)\): \((0,0,1,(q+1)(q^2+1))\). Every nonzero entry has column \(\ge_{\mathrm{dom}}\) row-label, as (V) predicts; \(m=2\) data of node 5 likewise. (Hand counts for \(m=3\): proved by the same block-by-block integrality analysis; not machine-checked.)

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