Proposition 3.7 · Disconnected stabilizers: (IG)$_n$ for all $n$ gives $\mathcal M_I=\mathrm{Irr}$

Disconnected stabilizers: (IG)$_n$ for all $n$ gives $\mathcal M_I=\mathrm{Irr}$

1. Points. Let $o\subseteq Y_{N,M}$ be a $\mathrm{Fr}$-stable orbit of $H=I/I_{N,M}$, $y\in o(\mathbb F_q)$ (Lang), $S=\mathrm{Stab}(y)$, $A=\pi_0S$ with the Frobenius automorphism $\sigma$, and $\Gamma=A\rtimes\langle\sigma\rangle$ ($\sigma$ of finite order). The $H(\mathbb F_q)$-orbits on $o(\mathbb F_q)$ are indexed by $H^1(\mathbb F_q,A)$, i.e. by the $A$-conjugacy classes in the coset $A\sigma\subseteq\Gamma$ (twisted classes $a\sim b\,a\,\sigma(b)^{-1}$): the point $hy$ with $h^{-1}\mathrm{Fr}(h)\in S$ goes to the image $a$ of $h^{-1}\mathrm{Fr}(h)$.
2. Sheaves on $o$. Equivariant irreducible local systems on $o$ are $\mathcal L_\xi$, $\xi\in\mathrm{Irr}(A)$, and $\mathrm{Fr}^*\mathcal L_\xi\simeq\mathcal L_{\xi\circ\sigma}$. If $\xi\circ\sigma\simeq\xi$, choose an extension $\tilde\xi$ of $\xi$ to $\Gamma$ (it exists since $\Gamma/A$ is cyclic [Isaacs, Cor. 11.22]); $\tilde\xi(\sigma)$ defines a Weil structure on $\mathcal L_\xi$, and identifying the stalk at $hy$ with $V_\xi$ through $h$, Frobenius acts by $\tilde\xi(a\sigma)$. So $\mathrm{tr}\,\mathcal L_\xi$ on $o(\mathbb F_q)$ is the function $a\mapsto\tilde\xi(a\sigma)$ on twisted classes.
3. Twisted characters are a basis. (i) Spanning: $\mathrm{Irr}(\Gamma)$ spans class functions on $\Gamma$; restricted to $A\sigma$ (where $\Gamma$-classes are $A$-classes, since $a^{-1}(a\sigma)a=\sigma(a)\sigma$), an irreducible $\chi$ of $\Gamma$ either restricts irreducibly to $A$, so $\chi=\tilde\xi\lambda$ with $\lambda\in\mathrm{Irr}(\Gamma/A)$ [Isaacs, Cor. 6.17 (Gallagher)], and $\chi|{A\sigma}=\lambda(\sigma)\tilde\xi|{A\sigma}$; or $\chi$ is induced from $A\rtimes\langle\sigma^e\rangle$, $e>1$ (Clifford), and vanishes on $A\sigma$. (ii) Orthonormality: for $\sigma$-stable $\xi,\eta$ put $s_j=|A|^{-1}\sum_{a\in A}\tilde\xi(a\sigma^j)\overline{\tilde\eta(a\sigma^j)}$; then $\langle\tilde\xi,\tilde\eta\lambda\rangle_\Gamma=d^{-1}\sum_j\overline{\lambda(\sigma^j)}s_j$ equals $\delta_{\xi\eta}\delta_{\lambda1}$ (Gallagher: the $\tilde\xi\lambda$ are distinct), so by Fourier inversion on $\mathbb Z/d$, $s_1=\delta_{\xi\eta}$. Hence $\{\tilde\xi|{A\sigma}\}{\xi\ \sigma\text{-stable}}$ is a basis of the $A$-class functions on $A\sigma$.
4. Basis at a level. With step 3, the traces of the $\mathrm{Fr}$-stable $\mathrm{IC}(P,\xi)$ ($P$ over the $\mathrm{Fr}$-stable pieces) are block-triangular in the indicators of the $I(\mathcal O)$-orbits of $C_I^{(N,M)}$, with invertible diagonal blocks; non-$\mathrm{Fr}$-stable pieces have no rational points. So they form a basis of $C_I^{(N,M)}\otimes\overline{\mathbb Q}_\ell$, and Lemma 3.4(i)–(iii) hold with $\mathrm{Irr}$ replaced by the $\mathrm{Fr}$-stable simple objects. The chain of Theorem 3.6 gives (a).
5. (b). Frobenius permutes the finitely many simple objects of a level, so each $S$ is $\mathrm{Fr}^n$-stable for some $n$. Replacing $q$ by $q^n$ changes neither $\mathcal C_I$ nor $\mathcal M_I$ (both live over $\overline{\mathbb F}q$), so (a) over $\mathbb F{q^n}$ puts $S$ in $\mathcal M_I$. Conclude with Lemma 3.5(b). $\square$

Where this is needed. The symmetric rows of Table 1 whose Iwahori stabilizers at finite level are disconnected (for instance through $\pi_0Z_{G^\theta}(\cdot)$); for $\mathbb A^n$ and the group stabilizers are connected (Corollary 3.8, Proposition 3.9).