GL_1 and the trivial parabolic endpoints

At $n=1$ the three defining conditions of [F1] read: a $1\times 1$ matrix is upper triangular and diagonal always, and it is upper unitriangular exactly when its entry is $1$; hence $B=T=G$ and $U=\{I_1\}$. By [F2] the monomial matrices are all of $G$, so $N=G$, $W=N\mathrm{/}T$ is the trivial group, and it is $S_1=\{1\}$ under the isomorphism $\sigma \mapsto P_{\sigma }T$.

For the composition $\alpha =(1)$ of $n=1$ the block is $I_1=\{1\}$, so ${\mathrm{blk}}_{\alpha }(1)={\mathrm{blk}}_{\alpha }(1)$ and the criteria of [F2] impose no vanishing condition: $P_{\alpha }=L_{\alpha }=G$; the condition for $U_{\alpha }$ is $g_{11}=1$, so $U_{\alpha }=\{I_1\}$. In particular $P_{\alpha }=G=L_{\alpha }$ and $U_{\alpha }=\{I_1\}$, and the standard parabolic is the whole group.

The set $\{1\}$ has exactly one ordered partition, namely ${\gamma }_0=(\{1\})$, whose single block is $\{1\}$; a proper refinement of ${\gamma }_0$ would be an ordered partition of $\{1\}$ whose set partition differs from $\{\{1\}\}$, and no such partition exists, since an ordered partition of $\{1\}$ consists of nonempty disjoint blocks covering $\{1\}$. Hence ${\gamma }_0$ has no proper refinement at all.

The vectors of $V={\mathbb{F}}_q^1$ are the scalar multiples of $e_1$, so the only subspace different from $0$ is $V$ itself and the chain $0<V$ is the only complete flag of $V$; by [F3] the coset space $G\mathrm{/}B$ is a single point, and by $B=T=G$ of step 1.1 the standard flag is fixed by every element of $G$.

The inflation ${\mathrm{Inf}}_{L_{\alpha }}^{P_{\alpha }}$ of [F5] is taken along the identity homomorphism $P_{\alpha }\to L_{\alpha }$, since $P_{\alpha }=L_{\alpha }=G$, so it is the identity functor on complex $G$-modules; likewise ${\mathrm{Ind}}_{P_{\alpha }}^G$ is the identity functor, because $P_{\alpha }=G$. Hence $R_G^G(W)=W$ for every complex $G$-module $W$, and ${}^{\ast }\text{\hspace{0.17em}⁣}{R}_G^G(X)=X^{U_{\alpha }}=X^{\{I_1\}}=X$ for every complex $G$-module $X$: both Harish-Chandra functors attached to the one-part composition of $n=1$ are the identity functors.

By [F4] the group is the disjoint union of the cells $B{P}_{\sigma }B$ over $\sigma \in S_1=\{1\}$, so $G=B{P}_{\mathrm{i}\mathrm{d}}B=B$ by step 1.1: there is exactly one Bruhat cell, namely the cell of the identity, and it is the single right coset of $B=G$ in $G$; this matches $\mathrm{\mid }B{P}_{\mathrm{i}\mathrm{d}}B\mathrm{/}B\mathrm{\mid }=q^{\mathrm{\ell }(\mathrm{i}\mathrm{d})}=q^0=1$ of [F4] and the single point $G\mathrm{/}B$ of step 2.1.

Let $M$ be a complex $G$-module. By step 1.3 the only ordered partition of $\{1\}$ is ${\gamma }_0$, which has no proper refinement, so the vanishing condition of [F6] is vacuous and $M$ is cuspidal. If moreover $M=\chi$ is simple, then $(G,\chi )$ is a cuspidal pair, and its Harish-Chandra series consists of the simple quotients of $R_G^G(\chi )=\chi$ (step 2.2), that is of $\chi$ itself: the series is the singleton $\{\chi \}$. Every simple complex $G$-module is one-dimensional, since $G={\mathbb{F}}_q^{\times }$ is abelian, and each of them forms its own Harish-Chandra series.

Consequently, at $n=1$ the general theory takes the following form: $B=T=G$, $U=\{I_1\}$ and $W=S_1$; the complete flag variety $G\mathrm{/}B$ and the set of Bruhat cells and double cosets $B\mathrm{\backslash }G\mathrm{/}B$ each have exactly one element; the one-part composition $(1)$ has $P_{\alpha }=L_{\alpha }=G$ with trivial unipotent radical, so both Harish-Chandra functors are the identity; and every simple module is cuspidal with a one-element series. ∎