Audit 5.6 · Audit of A3-6: orbits \((d_1,d_2,e)\), connected stabilizers, \(\delta=3d_1+3d_2+e\)

Re-derivation.

1. One plane. \(N=W\cap L\) has dual \(N^\#=\mathrm{pr}_W(L)\), because \(L=L^\#\). So \(L/(N\oplus N')\) is the graph of an anti-isometry \(\varepsilon:N^\#/N\to N'^\#/N'\), with \(N^\#/N\cong(\mathcal O/t^d)^2\) and the same \(d\) for \(N'\). Two gluings differ by an element of \(\mathrm{Sp}(N'^\#/N')=\mathrm{SL}_2(\mathcal O/t^d)\), which lifts to \(\mathrm{SL}(N')\). So planes of the same type \(d\) are \(K\)-conjugate, and \(\operatorname{Stab}(W)\) surjects onto \(\mathrm{SL}(N)\).
2. Isometry. In bases with \(\omega=t^{d_i}\), \(\det\varphi=t^{d_1-d_2}\) exactly. Absorbing \(\operatorname{diag}(u,1)\) shows that \(\mathrm{SL}_2(\mathcal O)\)-double cosets with fixed determinant are given by the Smith form. Hence the labels \((d_1,d_2,e)\), \(e\equiv d_1-d_2\pmod 2\), and \(X(\mathcal O)\) is the label \((0,0,0)\).
3. Stabilizers. \(a_1\in\Gamma_0(t^e)=\mathrm{SL}(N_1)\cap\mathrm{SL}(\varphi^{-1}N_2)\) is connected: it is an extension of the Borel subgroup of \(\mathrm{SL}_2\) by a pro-unipotent group when \(e\ge1\), and \(\mathrm{SL}_2(\mathcal O)\) when \(e=0\). Over each \(a_1\), the \(b_i\) form a torsor under the level-\(d_i\) congruence subgroup (connected). So \(\operatorname{Stab}_{L^+G}(x_\nu)\) is connected, and Lang's theorem gives one \(K\)-orbit per label. This agrees with step 2 run over \(\mathbb F_q\).
4. Dimension. With invariant measures normalized by \(dg=dx\,dh\), \(\operatorname{vol}(Kgx_0)=\operatorname{vol}(K)/\operatorname{vol}_H(g^{-1}Kg\cap H(F))\). Transport by \(g\), and note that all hyperspecial subgroups \(\mathrm{SL}(M)\) of \(\mathrm{SL}(W)(F)\) are \(\mathrm{GL}(W)\)-conjugate, and that \(\mathrm{GL}(W)\)-conjugation preserves Haar measure on \(\mathrm{SL}(W)(F)\). This gives the displayed index. Its leading exponent is \(e+3d_1+3d_2\), which is A3-6(c).
5. Finiteness. \(\mathcal C_\lambda=m(\mathrm{Gr}_{\le\lambda}\tilde\times L^+X)\) is closed: \(LG\times^{L^+G}L^+X\hookrightarrow\mathrm{Gr}\times LX\) is closed because \(L^+X\subset LX\) is closed (\(X\) affine), and \(\mathrm{Gr}_{\le\lambda}\) is proper. If \(x\) is integral for \(L'\) of position \(\le\lambda\), then \(t^{a_i}C\subseteq N_i\) for the unimodular \(C=W_i\cap L_i'\). This forces \(\omega(N_i,N_i)\supseteq t^{2a_i}\mathcal O\), so \(d_i\le2a_i\), and \(e\le D^1+D^2\le2(a_1+a_2)\) by the triangle inequality. So each \(\mathcal C_\lambda\) contains finitely many orbits. \(LX(k)=LG(k)x_0\), so every orbit closure lies in some \(\mathcal C_\lambda\).
6. Covered labels. \(\varphi(\lambda)=(a_1+b_1,a_2+b_2,(a_1-b_1)+(a_2-b_2))\) and \(\delta(\varphi(\lambda))=\langle2\rho,\lambda\rangle\). The image of \(\varphi\) is \(\{(d_1\bmod2)+(d_2\bmod2)\le e\le d_1+d_2\}\). The examples \((0,0,2k)\), \((1,1,0)\), \((1,0,3)\) are uncovered. Correct.

Supplement. The closure order is filtered. \(\mathcal C_\lambda\) is irreducible (the image of an irreducible space) and has finitely many orbits, so it is the closure of one orbit. Moreover \(\lambda\le\lambda+\lambda'\), because the coweight lattice of the simply connected \(\mathrm{Sp}_4^2\) equals the coroot lattice.

Remaining point. Antisymmetry of "\(\mathcal O_\mu\subseteq\overline{\mathcal O_\nu}\)" requires \(\delta\) to drop strictly on boundaries. That is automatic once the dimension theory of Objection 5.10 exists. It is not proved by A3-6.