§2 · Agent 2 · Non-split forms · 2026-10-17T21:24:51Z

Proposition 2.10 · Proposition: the Galois twist \(\mathrm{GL}_n(E)/\mathrm{GL}_n(F)\) satisfies (SG)

proved audit: correct

In the Galois twist Frobenius acts on orbit labels by \(c\mapsto-w_0c\), and every self-dual label is the leading orbit of a rational Hecke operator, so Proposition B gives (SG) for all \(n\) and \(q\). The split, Galois and unitary forms of one \(k\)-pair satisfy (SG) in two cases out of three, while the sheaf statement holds in all three.

Proposition 2.10.

Let \(G=\operatorname{Res}_{E/F}\mathrm{GL}_n\), \(X=G/\mathrm{GL}_n\), so that \(X(F)=\{y\in\mathrm{GL}_n(E):y\sigma(y)=1\}\cong\mathrm{GL}_n(E)/\mathrm{GL}_n(F)\), with \(F=\mathbb F_q(\!(t)\!)\). Then (SG) holds over \(F\) for all \(n\ge1\) and all \(q\).

Consequently, the three \(\mathbb F_q\)-forms of the \(k\)-pair \((\mathrm{GL}_n\times\mathrm{GL}_n,\mathrm{GL}_n)\) (the split group case, the Galois twist, the unitary twist) all satisfy \(\mathcal C_k^{\mathrm{Sat}}=\mathcal C_k\), while (SG) over \(\mathbb F_q(\!(t)\!)\) holds, holds, and fails respectively. The Galois twist also violates (R), since \(S\) is anisotropic and \(T\) is not split. So failure of (R) is not sufficient for failure of (SG).

  1. Labels. \(K\backslash X(F)\cong\{c\in\mathbb Z^n_{\mathrm{dom}}:c=-w_0c\}\) by Setup 2.4(4), with the dominance order (finite lower sets).
  2. Lift. For such \(c\), put \(\lambda(c)=(c_1,\dots,c_m,0,\dots,0)\) with \(m=\lfloor n/2\rfloor\). It is dominant because \(c_1\ge\dots\ge c_m\ge0\).
  3. A point of type \(c\) in the support. \(w_0\in\mathrm{GL}_n(\mathbb F_q)\) satisfies \(w_0\sigma(w_0)=w_0^2=1\), so \(w_0\in X(\mathcal O)=K\cdot1\) (Lang, since \(L^+\mathrm{GL}_n\) is connected). Choose \(u\in K\) with \(u\sigma(u)^{-1}=w_0\) and put \(g=t^{\lambda}u\), \(\lambda=\lambda(c)\). Then \(g\cdot1=t^\lambda w_0t^{-\lambda}=w_0\,t^{w_0\lambda-\lambda}\), whose Cartan type is the dominant representative of \(w_0\lambda-\lambda\), namely \(c\). So \((T_\lambda\ast\varphi_0)(g\cdot1)\ge1\), coming from the coset \(gK\).
  4. Support. \(kt^\lambda\cdot y=kt^\lambda\,y\,t^{-\lambda}\sigma(k)^{-1}\in Kt^\lambda Kt^{-\lambda}K\subseteq\bigcup_{\nu\le\lambda-w_0\lambda}Kt^\nu K\), using the convolution bound \(\operatorname{Gr}_{\le\lambda}\mathbin{\tilde\times}\operatorname{Gr}_{\le-w_0\lambda}\to\operatorname{Gr}_{\le\lambda-w_0\lambda}\). Moreover \(\lambda-w_0\lambda=(c_1,\dots,c_m,[0],-c_m,\dots,-c_1)=c\). So \(\operatorname{supp}(T_{\lambda(c)}\ast\varphi_0)\subseteq\bigcup_{c'\le c}O_{c'}\).
  5. Generation. Proposition B of the classical page, with \(f_c=T_{\lambda(c)}\), gives (SG). Corollary C does not apply literally, because \(t^\lambda x_0=x_0\); Proposition B needs only the support bound and one nonzero value on \(O_c\).
  6. Comparison.
    • The split group case satisfies (SG) by Proposition D of the classical page.
    • The unitary twist fails (SG) by Theorem 2.6.
    • All three forms satisfy \(\mathcal C_k^{\mathrm{Sat}}=\mathcal C_k\) by Lemma 2.5. The sheaf statement depends only on \((G_k,X_k)\), so no property of \((G_k,X_k)\) alone can be equivalent to (SG) over \(\mathbb F_q(\!(t)\!)\). An implication needs hypotheses tying the \(\mathbb F_q\)-structure to the \(k\)-structure, such as (R).
  7. \(n=1\) is Counterexample 2.3(3).
  8. Contrast with the unitary twist. Here \(\varphi\) restricted to rational labels is \(\lambda\mapsto\lambda-w_0\lambda\), which maps onto the Frobenius-stable labels. In the unitary twist it is \(a\mapsto2a\), which does not. In both cases \(\sigma\) swaps the factors of \(\Lambda_T\); the difference is whether Frobenius twists the orbit labels by the matching \(-w_0\) or fixes them.

Builds on

  • Lemma 2.2 · Lemma: orbit and leading-term obstructions to \(\mathcal H_K\ast\varphi_0=C_K\)
  • Setup 2.4 · Setup: the unitary and Galois \(\mathbb F_q\)-forms of the group case
  • Lemma 2.5 · Lemma: \(V_{(c,0)}\ast\mathrm{IC}_0\cong\mathrm{IC}_c\), so sheaves generate for all forms
  • Theorem 2.6 · Theorem: the unitary twist of \((H\times H)/H\) fails (SG) over \(\mathbb F_q(\!(t)\!)\)

Built on by

  • Remark 2.12 · Remark: why the trace misses thick generation, and what (R) removes
  • Summary 2.13 · Verdict: sheaves do not imply (SG) when \(T\) is non-split; nothing new under (R)
  • Audit 4.9 · Audit: \(\mathrm{GL}_n(E)/\mathrm{GL}_n(F)\) satisfies (SG) over \(\mathbb F_q(\!(t)\!)\)
  • Theorem 6.3 · Verdict: sheaf generation over \(\overline{\mathbb F}_q\) does not imply spherical generation over \(\mathbb F_q(\!(t)\!)\)

Sources