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For a rational parabolic subgroup $P=L⋉U$, use inflation of a group representation for a complex $L^F$-module across $P^F\to L^F$ and induce it to $G^F$. This is Harish-Chandra induction. Its adjoint, Harish-Chandra restriction, takes $U^F$-invariants. These are ordinary finite-group group representation functors and should not be confused with general Deligne-Lusztig induction from a nonrational Borel.

For the special linear group $G={\mathrm{SL}}_3({\overline{\mathbb{F}}}_q)$, let $N=q-1$, and choose a generator $\chi$ of the character group of ${\mathbb{F}}_q^{\times }$. Every linear character of the diagonal rational maximal torus has a unique expressionThe upper triangular Borel subgroup $B=TU$ is rational. Consequently its Deligne-Lusztig induction is the actual principal series of a finite reductive groupwhere $\tilde{\theta }$ is the inflation of a group representation across $B^F\to T^F$. The geometric reason for this split case is $X(1)=G^F/B^F$: the associated torus covering supplies $\theta$, and the auxiliary unipotent fibres contribute a single even cohomological degree, hence no minus sign.

The rational Bruhat decomposition indexes $B^F\backslash G^F/B^F$ by $W=S_3$. Applying the Mackey restriction formula and Frobenius reciprocity givesEach intersection contains $T^F$, and the two linear characters are trivial on its unipotent part. Its contribution is therefore one exactly when $w$ fixes $\theta$, and zero otherwise. Complete reducibility over $\mathbb{C}$ makes this character inner product the sum of squares of irreducible representation multiplicities. Thus the principal series of a finite reductive group is irreducible exactly when the Weyl group stabilizer is trivial.

To compute that stabilizer, represent $\theta$ by the triple $({\chi }^r,{\chi }^s,1)$. Two triples give the same linear character on determinant-one diagonal matrices exactly when they differ by a common factor $(\nu ,\nu ,\nu )$. For example, triviality of $({\eta }_1,{\eta }_2,{\eta }_3)$ on all such matrices first gives ${\eta }_1={\eta }_3$ by varying $a$, and then ${\eta }_2={\eta }_3$ by varying $b$. The Weyl group permutes the three entries.

A transposition fixes $\theta$ precisely when its two interchanged entries coincide: the third, fixed entry forces the common factor to be one. These three obstructions are $r=0$, $s=0$, and $r=s$. A three-cycle can also fix $\theta$ without any two entries coinciding. Its common factor must satisfy ${\nu }^3=1$; in the nontrivial case the entries, up to order and common factor, are $(1,\nu ,{\nu }^2)$. This happens exactly when $3\mid N$ and $\{r,s\}=\{N/3,2N/3\}$.

The complete condition is thereforeIn particular pairwise distinct coordinate characters alone are insufficient. For $q=4$ the only distinct triples have the order-three symmetry, and no character of this split rational maximal torus gives an irreducible principal series of a finite reductive group.

Let $P=L⋉U$ and $Q=M⋉V$ be rational parabolic subgroups with rational Levi subgroups. Work with complex representations of their finite groups of Lie type. Here Harish-Chandra induction and Harish-Chandra restriction meanThe latter is an $M^F$-module, and it is adjoint to the former by Frobenius reciprocity.

DefineFor $x\in \mathcal{S}(M,L{)}^F$, put $H_x=M\cap {}^{x}L$ and let $\mathrm{ad}x$ transport an $L^F$-module to a $({}^{x}L{)}^F$-module. The Mackey formula for Harish-Chandra induction is the natural functor isomorphismOn characters the direct sum becomes a sum. The intersection parabolics displayed in the subscripts are important: they specify exactly which Harish-Chandra induction and Harish-Chandra restriction occur.

We use the following parabolic-intersection lemmas, stating explicitly the geometric input to the proof. First, representatives of $M^F\backslash \mathcal{S}(M,L{)}^F/L^F$ represent exactly the double cosets $Q^F\backslash G^F/P^F$. Second, for such $x$, the groups $M\cap {}^{x}P$ and $Q\cap {}^{x}L$ are parabolic subgroups of $M$ and ${}^{x}L$, respectively, with common Levi subgroup $H_x$ and respective unipotent radicals $M\cap {}^{x}U$ and $V\cap {}^{x}L$. If $K=Q\cap {}^{x}P$, projection $Q\to M$ maps $K$ onto $M\cap {}^{x}P$ with kernel $K\cap V$. Projection ${}^{x}P\to {}^{x}L$ maps $K\cap V$ onto $V\cap {}^{x}L$ with kernel $V\cap {}^{x}U$. These assertions hold on rational points as well. The surjectivity on rational points uses connectedness of the unipotent kernels and the Lang theorem for algebraic groups. These are the compatible-Levi and rational double-coset lemmas; they follow by applying the root-subgroup decomposition after choosing the common maximal algebraic torus.

For completeness, the representation-theoretic reduction is as follows. The group algebra realization of induced representation splits according to the double cosets:as a $(Q^F,P^F)$-bimodule. Tensoring on the right with an inflated $L^F$-module $E$ identifies each summand with induction from $K^F=(Q\cap {}^{x}P{)}^F$. This proves the finite-group Mackey restriction formula in the formHere ${}^x\tilde{E}$ is trivial on $({}^{x}U{)}^F$.

We next record the elementary normal-subgroup identity underlying Harish-Chandra restriction. If $N◃D$ are finite groups, with $N$ a normal subgroup, and $C\subset D$, then for a complex $C$-module $A$,To see this, replace invariants by coinvariants using the averaging idempotent $|N{|}^{-1}{\sum }_{n\in N}n$. Taking coinvariants in $\mathbb{C}[D]{\otimes }_{\mathbb{C}[C]}A$ first replaces $\mathbb{C}[D]$ by $\mathbb{C}[D/N]$, and forces $C\cap N$ to act trivially on $A$. Thus it yields $\mathbb{C}[D/N]{\otimes }_{\mathbb{C}[CN/N]}{A}_{C\cap N}$; averaging over $C\cap N$ identifies this last module with the one displayed. This also shows why characteristic zero makes the argument exact.

Apply this identity to $D=Q^F$, $N=V^F$, $C=K^F$. By the intersection lemmas, $CN/N=(M\cap {}^{x}P{)}^F$. Since ${}^x\tilde{E}$ is trivial on $({}^{x}U{)}^F$, its $(K\cap V{)}^F$-invariants are exactlyMoreover $(M\cap {}^{x}U{)}^F$ acts trivially on this module, so induction from $(M\cap {}^{x}P{)}^F$ to $M^F$ is precisely Harish-Chandra induction from $H_x$. Substituting into each double-coset summand gives the boxed formula. This proves the formula for rational parabolics; it does not assume a general Mackey formula for nonrational Deligne-Lusztig induction.

Write $\chi =\epsilon R_T^G(\theta )$, where $\epsilon \in \{1,-1\}$ and $\chi$ is the character of an irreducible representation. An irreducible representation is a cuspidal representation of a finite reductive group exactly when its Harish-Chandra restriction to every proper rational Levi subgroup of a rational parabolic subgroup is zero.

The needed identity is the Harish-Chandra restriction of a Deligne-Lusztig character. For a rational parabolic subgroup $P=M⋉V$, the torus case of the induction-restriction formula givesHere ${}^x\theta (xt{x}^{-1})=\theta (t)$. The geometric meaning of the indexing condition is that only a rational conjugate of the inducing maximal algebraic torus lying in the Levi can survive taking $V^F$-invariants. This is the established torus case of Deligne-Lusztig induction, valid without a restriction on $q$; it is not an appeal to an unrestricted induction-restriction conjecture. See Dudas and Michel's lectures, Theorem 14.4.

Suppose first that $T$ is an elliptic maximal torus, meaning it lies in no proper rational parabolic subgroup. For any proper rational $P=M⋉V$, the indexing set in the displayed sum is empty. Indeed ${}^{x}T\subset M$ with $x\in G^F$ would imply $T\subset {}^{x^{-1}}P$, itself a proper rational parabolic subgroup. Hence ${}^{\ast }{R}_{M\subset P}^G{R}_T^G(\theta )=0$, and multiplication by $\epsilon$ gives ${}^{\ast }{R}_{M\subset P}^G\chi =0$. This proves cuspidality.

Conversely suppose $T\subset P$ for a proper rational parabolic subgroup. There is a unique Levi subgroup of $P$ containing this particular maximal algebraic torus: relative to $T$, it is generated by $T$ and those root subgroups for which both the root and its negative occur in $P$. Call it $M$. Uniqueness and $F(T)=T$, $F(P)=P$ imply $F(M)=M$, so this is a proper rational Levi. Transitivity of Deligne-Lusztig induction through this rational parabolic saysSet $\psi =R_T^M(\theta )$, allowing it to be a virtual character. Adjunction of Harish-Chandra induction and Harish-Chandra restriction, extended linearly to virtual characters, givesThus this Harish-Chandra restriction cannot vanish, so $\chi$ is not cuspidal. No positivity or irreducibility of $\psi$ is needed, and the sign in $\chi$ causes no cancellation problem. We have provedThe term elliptic maximal torus is relative to the ambient group: split central factors are allowed. The actual condition is absence of a proper rational parabolic subgroup containing $T$.