Audit 5.2 · Audit of A3-3: the model of X, its stabilizer, the dual group and fusion

Re-derivation.

1. Stabilizer. \((g_1,g_2)\) fixes \((W_1^0,W_2^0,\varphi_0)\) iff \(g_iW_i^0=W_i^0\) and \(g_2\varphi_0=\varphi_0g_1\) on \(W_1^0\). Since \(\operatorname{Stab}(W_i^0)=\mathrm{Sp}(W_i^0)\times\mathrm{Sp}(W_i^{0\perp})\), this gives \(H=\mathrm{Sp}(W_1^{0\perp})\times\mathrm{Sp}(W_2^{0\perp})\times\{(g,\varphi_0g\varphi_0^{-1})\}\cong\mathrm{SL}_2^3\) and \(\dim X=20-9=11\). Sakellaridis's text (sak.txt, Example 8.0.1) reads: "the first and second copies of \(\mathrm{SL}_2\) are embedded as \(\mathrm{Sp}_2\) in the first and second copies of \(\mathrm{Sp}_4\), while the third copy of \(\mathrm{SL}_2\) is embedded diagonally in both". This matches.
2. Points. \(H^1(E,\mathrm{SL}_2^3)=1\) for every field \(E\), so \(X(E)=G(E)/H(E)\). The graph model is correct. If \(\Gamma\) is isotropic for \(\omega_1-\omega_2\) and \(\omega_1|_{\mathrm{pr}_1\Gamma}\ne0\), then both projections are injective onto non-degenerate planes, and \(\Gamma\) is the graph of an isometry. The complement of \(X\) is a hyperplane section, so \(X\) is affine; also \(\dim\mathrm{IG}(2,8)=11\).
3. Sphericity and rank (independent check). A generic \(W_i\) has stabilizer in \(B_i\) equal to the Borel \(B^\Delta\) of the twisted diagonal \(\mathrm{SL}_2\), and it acts on \(W_i\) through a Borel \(B_i'\) of \(\mathrm{SL}(W_i)\). Then \(B_2'\times B_1'\) acts on \(\operatorname{Isom}(W_1,W_2)\cong\mathrm{SL}_2\) by \(\varphi\mapsto b_2\varphi b_1^{-1}\), with an open (big-cell) orbit. So \(X\) has an open \(B\)-orbit with generic stabilizer of dimension 1 (a torus). Since the three spherical roots involve all four simple roots, \(\dim X=\dim U+\operatorname{rk}X\) gives rank 3.
4. Dual group. The text gives \(\mathrm{SL}_2\times\mathrm{SL}_2^{\rm diag}\times\mathrm{SL}_2\to\mathrm{SL}_2^4\to\mathrm{SO}_4\times\mathrm{SO}_4\to\mathrm{SO}_5\times\mathrm{SO}_5\), "where diag denotes the diagonal embedding into the second and third copies", and \(\delta(X)^{1/2}\) is trivial. This is A3-3(c). \(\ker\iota=\{\pm(1,1,1)\}\) is correct.
5. Fusion. \(\iota(a,b,c)=((ab,a/b),(bc,b/c))\). Inverting the second coordinate in both factors (an element of \(W(B_2)^2\)) sends \(\iota(a,b,a)\) to \(\iota(b,a,b)\). In \(\check G_X=\mathrm{SL}_2^3/\{\pm1\}\), two torus points are conjugate iff \((a',b',c')=\epsilon\,(a^{\pm1},b^{\pm1},c^{\pm1})\) with \(\epsilon=\pm1\). For \((a,b,a)\) and \((b,a,b)\) this forces \(b=\epsilon a^{\pm1}\). So non-conjugacy needs \(a\ne\pm b^{\pm1}\), not just \(a\ne b^{\pm1}\). Counterexample to (d) as written: \(a=-b\) with \(b^4\ne1\), where \(\iota(-b,b,-b)=\iota(b,-b,b)\).
6. The defect is harmless downstream. Audit 5.9 uses (d) only at generic \((a,b)\), and the fused locus is still the divisor \(\{c=a\}\).

Remark (relevant to A3-10). \(X\) is a closed \(G\)-orbit in a symmetric variety. Put \(Y'=\{\Gamma\subset V_1\oplus V_2:\ (\omega_1+\omega_2)|_\Gamma\ \text{non-degenerate}\}\cong\mathrm{Sp}_8/(\mathrm{Sp}_6\times\mathrm{Sp}_2)\). For \(p\ne2\), \(X=\{\Gamma\in Y':\Gamma\ \text{isotropic for}\ \omega_1-\omega_2\}\), since on such \(\Gamma\) we have \((\omega_1+\omega_2)|_\Gamma=2\,\omega_1\circ\mathrm{pr}_1\). Equivalently, \(H=G\cap\mathrm{Sp}_8^{s}\) for the symplectic involution \(s\) that swaps \(W_1^0\) and \(W_2^0\) through \(\varphi_0\) and fixes \(P_1\oplus P_2\).