Correction 10.6 · Correction: the \(m=3\) check table for \(\GL_6/\Sp_6\), with the Lagrangians that are graphs

1. Self-dual lattices as Lagrangians. By Computation 5.5, \(N(\lambda,b')\) counts the lattices \(\Lambda''\supseteq\Lambda_0\) of type \(\lambda\) on which \(x_{b'}\) is unimodular. Such a lattice is self-dual, so \(\Lambda_0\subseteq\Lambda''\subseteq\Lambda_0^\#\) as in the proof of Lemma 3 of Lemma 5.6, and \(\Lambda''\mapsto\Lambda''/\Lambda_0\) is a bijection onto the Lagrangian submodules of the finite symplectic module \(M'_{b'}=\Lambda_0^\#/\Lambda_0=\bigoplus_i(\cO/\varpi^{b'_i})^2\), with the \(F/\cO\)-valued form \(\langle u,v\rangle=\sum_i\varpi^{-b'_i}\det(u_i,v_i)\). So \(N(\lambda,b')\) is the number of Lagrangian submodules of \(M'_{b'}\) of module type \(\lambda\).

2. Lagrangians of an orthogonal sum. Let \(M'=\mathcal A\perp\mathcal C_K\). For a Lagrangian \(M\) put \(K_{\mathcal A}=M\cap\mathcal A\) and \(K_{\mathcal C_K}=M\cap\mathcal C_K\). Isotropy puts the projections of \(M\) inside \(K_{\mathcal A}^\perp\) and \(K_{\mathcal C_K}^\perp\), and comparing \(|M|^2=|M'|\) with \(|K^\perp|=|\mathcal A|/|K|\) forces equality; so \(M\) is the graph of an anti-isometry \(K_{\mathcal A}^\perp/K_{\mathcal A}\to K_{\mathcal C_K}^\perp/K_{\mathcal C_K}\). Conversely, every pair of isotropic submodules together with such an anti-isometry gives exactly one Lagrangian. Between two nondegenerate symplectic planes over \(\F_q\) there are \(|\SL_2(\F_q)|=q(q^2-1)\) anti-isometries.

3. The column \(b'=(3,1,0)\). Here \(\mathcal A=(\cO/\varpi^3)^2\) and \(\mathcal C_K=(\cO/\varpi)^2=\F_q^2\); the third block is zero.
   • \(K_{\mathcal C_K}\) a line (\(q+1\) choices). Then \(M=K_{\mathcal A}\oplus K_{\mathcal C_K}\) with \(K_{\mathcal A}\) Lagrangian in \(\mathcal A\), and \(M\) has type \((3,1)\) iff \(K_{\mathcal A}=\cO u\) with \(u\) primitive, which gives \(|\mathbb P^1(\cO/\varpi^3)|=q^2(q+1)\) choices. The Lagrangians of \(\mathcal A\) of type \((2,1)\) are the \(q+1\) modules \(\varpi^2\mathcal A+\cO\varpi u\), and give type \((2,1,1)\). Contributions: \(q^2(q+1)^2\) to type \((3,1)\) and \((q+1)^2\) to type \((2,1,1)\).
   • \(K_{\mathcal C_K}=0\). Then \(K_{\mathcal A}^\perp/K_{\mathcal A}\cong\F_q^2\), so \(|K_{\mathcal A}|=q^2\). Either \(K_{\mathcal A}=\varpi^2\mathcal A\), with \(K_{\mathcal A}^\perp=\varpi\mathcal A\), and then \(M\subseteq\varpi\mathcal A\oplus\mathcal C_K\) has exponent \(2\) and \(|\varpi M|=q^2\), so type \((2,2)\); or \(K_{\mathcal A}=\cO\varpi u\) with \(u\) primitive modulo \(\varpi^2\) (\(q(q+1)\) choices), \(K_{\mathcal A}^\perp=\cO u+\varpi^2\cO u'\) where \(\det(u,u')=1\), and \(M\) contains an element \((u,c)\) of order \(\varpi^3\), so has type \((3,1)\). Each choice allows \(q(q^2-1)\) anti-isometries. Contributions: \(q(q^2-1)\) to type \((2,2)\) and \(q(q+1)\cdot q(q^2-1)=q^2(q+1)^2(q-1)\) to type \((3,1)\). Total for type \((3,1)\): \(q^2(q+1)^2\big(1+(q-1)\big)=q^3(q+1)^2\).

4. The column \(b'=(2,1,1)\). Here \(\mathcal A=(\cO/\varpi^2)^2\) and \(\mathcal C_K=\F_q^4\).
   • \(K_{\mathcal C_K}\) Lagrangian in \(\mathcal C_K\) (\((q+1)(q^2+1)\) choices) and \(M=K_{\mathcal A}\oplus K_{\mathcal C_K}\) with \(K_{\mathcal A}=\cO u\), \(u\) primitive (\(q(q+1)\) choices): \(q(q+1)^2(q^2+1)\) Lagrangians of type \((2,1,1)\).
   • \(K_{\mathcal C_K}=\ell\) a line (\((q+1)(q^2+1)\) choices), so \(K_{\mathcal C_K}^\perp/K_{\mathcal C_K}\cong\F_q^2\). Then \(|K_{\mathcal A}|=q\), \(K_{\mathcal A}=\cO\varpi u\) (\(q+1\) choices), \(K_{\mathcal A}^\perp=\cO u+\varpi\mathcal A\), and there are \(q(q^2-1)\) anti-isometries; since \(\varpi M=\varpi K_{\mathcal A}^\perp=\cO\varpi u\) has order \(q\), \(M\) has type \((2,1,1)\). Total: \((q+1)^2(q^2+1)\,q(q^2-1)\).
   • \(K_{\mathcal C_K}=0\) is impossible: it would force \(K_{\mathcal A}=0\) and \(\mathcal A\cong\F_q^4\). Total: \(q(q+1)^2(q^2+1)\big(1+(q^2-1)\big)=q^3(q+1)^2(q^2+1)\).

5. What went wrong, and what is unaffected. The values in Lemma 5.6 are exactly the first bullets of steps 3 and 4: the Lagrangians that split along \(\mathcal A\perp\mathcal C_K\). The Lagrangians that are graphs of anti-isometries were missed. All corrected entries are nonzero where Lemma 5.6 has nonzero entries, so the zero pattern, and with it the consistency check of Lemma 3, is unchanged; Lemmas 1–3 and Theorem 5.7, which uses only Lemmas 2 and 3, are unaffected. An exhaustive count of Lagrangian submodules over \(\F_2[[t]]\) and \(\F_3[[t]]\), and a direct enumeration of lattices in Hermite normal form over \(\F_2[[t]]\), give the table for \(q=2,3\).

Verified by computation: symp_lagr.py, check_V.py, symp_hnf.py.