§4 · Agent 4 · Audit: counterexample and mechanism · 2026-10-17T22:13:11Z

Audit 4.7 · Audit: \(G(F)\)-orbits of the hermitian spaces, and failure on the base orbit

proved

For hermitian \(\mathrm{GL}_n\) there are exactly two \(G(F)\)-orbits, separated by \(v(\det)\bmod2\); for \(n\ge2\) generation already fails on the base orbit, by the partition count \(p_{\le n}(2d)-p_{\le n}(d)\). For hermitian \(\mathrm{SL}_2\) of determinant \(1\) there is a single \(G(F)\)-orbit, so that failure is not an orbit obstruction.

Audit 4.7.

  1. For \(H=\mathrm{GL}_n\), \(n\ge1\), \(X_u(F)\) (nondegenerate hermitian matrices) has exactly two \(G(F)\)-orbits: \(G(F)x_0=\{v(\det x)\text{ even}\}\) and its complement.
  2. For \(n\ge2\) and \(d\ge1\), \(L_d=\{c\le(2d,0,\dots,0)\}\) lies in the base orbit, \(\#L_d=p_{\le n}(2d)\), and \(\mathcal H_K\ast\varphi_0\cap C(L_d)\) has codimension \(p_{\le n}(2d)-p_{\le n}(d)\ge1\) in \(C(L_d)\).
  3. For \(H=\mathrm{SL}_2\), \(X_u(F)=\{x\in\mathrm{SL}_2(E):x^*=x\}\) is a single \(G(F)\)-orbit, and \(\mathbf 1_{K\cdot\operatorname{diag}(t,t^{-1})}\notin\mathcal H_K\ast\varphi_0\).

Outcome: holds (Theorem 2.6(5), steps 6–7; Counterexample 1.5 step 8).

  1. Invariant. \(\det(gxg^*)=N(\det g)\det x\) and \(v(Ny)=2v(y)\) on \(E^\times\), so \(v(\det x)\bmod2\) is \(G(F)\)-invariant; it is \(0\) on \(X(\mathcal O)\).
  2. The even classes form one orbit. Every \(x\) is \(K\)-equivalent to some \(t^c\) (Audit 4.3). Scaling a diagonal entry by \(N(t^k)=t^{2k}\) changes \(c_i\) by \(2k\), and \(g'=\left(\begin{smallmatrix}a&1\\-1&\bar a\end{smallmatrix}\right)\) with \(Na=t-1\) (such \(a\in\mathcal O_E^\times\) exists because \(N(\mathcal O_E^\times)=\mathcal O^\times\)) satisfies \(g'g'^*=(Na+1)I_2=tI_2\). So each \(t^c\) with \(\sum c_i\) even lies in \(G(F)\cdot1\). Cohomologically: \(H^1(F,U_n)\cong F^\times/NE^\times\cong\mathbb Z/2\) and \(H^1(F,\operatorname{Res}\mathrm{GL}_n)=1\).
  3. Count. \(c\le(2d,0,\dots,0)\) means \(c\) dominant, \(\sum c_i=2d\) and all partial sums \(\le2d\); the last partial sum forces \(c_n\ge0\). So \(L_d\) is the set of partitions of \(2d\) into at most \(n\) parts, \(\#L_d=p_{\le n}(2d)\), and \(\#\{a:2a\in L_d\}=p_{\le n}(d)\). By Audit 4.5(3) the codimension is the difference; \((2d-1,1,0,\dots,0)\) is not of the form \(2a\). Confirmed numerically for \(n=2\), \(d=1,2,3\) (Computation 4.6).
  4. \(\mathrm{SL}_2\), one orbit. \(G^{\theta_u}(F)\) is the special unitary group of the hermitian form \(1_2\). That form is isotropic: \(Nu=-1\) has a solution \(u\in\mathcal O_E^\times\), and \(h((1,u),(1,u))=1+Nu=0\). So \(G^{\theta_u}\cong\mathrm{SL}_2\) over \(F\), \(H^1(F,G^{\theta_u})=1\), and there is one \(G(F)\)-orbit. The packet's own argument also checks out: a binary hermitian form is a \(4\)-dimensional quadratic form over \(F\), \(p\) odd, which represents every element of \(F^\times\); \(\operatorname{diag}(t,t^{-1})\) represents \(1\) at \((t^{-1},v_0)\) with \(Nv_0=t-1\), because \(t\cdot t^{-2}+(t-1)t^{-1}=1\); and the isometry can be moved into \(\mathrm{SL}_2(E)\) by \(\operatorname{diag}(u,1)\), \(Nu=1\).
  5. Non-generation for \(\mathrm{SL}_2\). The labels are \(c\alpha^\vee\), \(c\ge0\), with the usual order on \(\mathbb Z\) (\(\Lambda=\mathbb Z\alpha^\vee\)); the leading labels are \(2a\). On \(L=\{0,1,2\}\) the span is that of \((1,0,0)\) and \((q(q+1),q+1,1)\) (Computation 4.6), which does not contain \((0,1,0)\).
  6. Counterexample 1.5 step 8 (unitary \(\mathrm{GL}_m\)): \(\mathbf 1_{K\cdot\operatorname{diag}(t,1,\dots,1)}\) lies in the odd orbit, outside \(\operatorname{supp}(\mathcal H_K\ast\varphi_0)\subseteq G(F)x_0\): correct.

Builds on

  • Theorem 2.6 · Theorem: the unitary twist of \((H\times H)/H\) fails (SG) over \(\mathbb F_q(\!(t)\!)\)
  • Counterexample 1.5 · Counterexample outside (R): the norm torus quotient, and the torus criterion

Built on by

  • Summary 4.12 · Verdict of the audit: the unitary twist is a valid counterexample; the repairs are minor

Sources