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For a rational maximal torus $T$, choose a Borel subgroup containing it with unipotent radical $U$. The alternating compactly supported étale cohomology of ${\mathcal{L}}_F^{-1}(U)$, in the appropriate $T^F$-isotypic component, defines the virtual character of $G^F$ $R_T^G(\theta )$. It is independent of the chosen Borel. When the Borel subgroup is rational, this is ordinary induction from its rational points after inflation of a group representation from $T^F$.

A SU(2) representation descends to the SO(3) group exactly when $-I$ acts trivially. In ${\mathrm{Sym}}^n({\mathbb{C}}^2)$ it acts by $(-1{)}^n$, so descent holds precisely for even $n$, or integer spin $j=n/2$. This is the inflation of a group representation criterion for the kernel $\{\pm I\}$. Half-integer-spin representations instead remain representations of the covering group.

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.

For a representation $\rho$ of $G/N$, its inflation of a group representation to $G$ is $g\mapsto \rho (gN)$, with character $\tilde{\chi }(g)=\chi (gN)$. Inflation preserves irreducibility because the invariant subspaces do not change.

For irreducible complex characters ${\chi }_i$, the row character orthogonality areThe column relations are ${\sum }_i{\chi }_i(g)\overline{{\chi }_i(h)}=0$ for nonconjugate $g,h$, and $|C_G(g)|$ when $g,h$ are conjugate. Class sizes convert the row sum into its usual weighted character-table form.

The commutator subgroup, or derived subgroup, isA linear character is afforded by a one-dimensional representation $\rho :G\to {\mathbb{C}}^{\times }$. Since ${\mathbb{C}}^{\times }$ is abelian, every commutator lies in $\ker \rho$, so

Let $\pi :G\to G/G^{\prime }$ be the quotient map. Every irreducible representation of a finite abelian group is one-dimensional, so each irreducible character $\psi$ of the abelianization $G/G^{\prime }$ inflates to the linear character $\psi \circ \pi$ of $G$. Conversely, the containment $G^{\prime }\le \ker \chi$ makes every linear character of $G$ factor uniquely through $G/G^{\prime }$. Thus inflation of a group representation gives a bijection