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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.