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This is the Bruhat decomposition associated with a BN-pair. The Bruhat decomposition is the disjoint unionassociated with a BN-pair.

This is the finite special-linear-group instance of the general Bruhat decomposition.

For the upper-triangular subgroup $B$ of $S{L}_2({\mathbb{F}}_p)$ andthe two double cosets are $B$ and $BwB$, and $B\cap wB{w}^{-1}=T$, the diagonal subgroup.

Use unnormalized parabolic induction, the convention fixed by the supplied sequence and by the occurrence of constants in $V(1)$. Its functions satisfy $f(bg)=\chi (b)f(g)$, and $G$ acts by right translation. Characters in this smooth category are smooth characters. Put $\lambda ={\chi }^w{\delta }^{-1}$, retaining the question's modulus $\delta (\mathrm{diag}(a,d))=|d/a|$.

First derive the evaluation adjunction for unnormalized induction. A $G$-intertwiner $A:V(\chi )\to V(\xi )$ gives a functional $\ell (v)=A(v)(1)$. Since $A$ intertwines and induced functions have left covariance, $\ell (\pi (b)v)=\xi (b)\ell (v)$. Conversely such a functional definesThis function has the required left covariance. Any compact-open subgroup fixing $v$ fixes it on the right, so it is a smooth induced function; compactness of $B\backslash G$ removes any additional support issue. The construction is $G$-equivariant, and the two operations are inverse. Since $\xi$ is trivial on $U$, these functionals factor precisely through the Jacquet module, givingNow use the given exact sequence $0\to \lambda \to J\to \chi \to 0$, where $J=V(\chi {)}_U$ has dimension two.

If $\lambda \ne \chi$, choose $t_0\in T$ on which their values differ. Its operator on $J$ is upper triangular with distinct eigenvalues, so has two one-dimensional eigenspaces. Since $T$ is abelian, all its operators commute with this operator and preserve the eigenspaces. They are the linear character lines $\lambda$ and $\chi$. This proves that distinct-character extensions of an abelian group split, so $J\cong \lambda \oplus \chi$. A map between one-dimensional linear character spaces is zero unless the linear characters agree, and otherwise is one-dimensional. Therefore the required dimension is one for $\xi =\lambda$ or $\xi =\chi$, and zero for any other linear character.

The case $\lambda =\chi$ needs an extra argument: the exact sequence alone would also permit a split self-extension with two quotient functionals. Write $\chi (\mathrm{diag}(a,d))={\chi }_1(a){\chi }_2(d)$. Equality with $\lambda$ forces ${\chi }_1/{\chi }_2=|\cdot |$, so $\chi =(\eta \circ \det ){\chi }_0$, where $\eta ={\chi }_2$ and ${\chi }_0(\mathrm{diag}(a,d))=|a|$. Multiplication of induced functions by $\eta (\det g)$ identifies this with a determinant-character twist of $V({\chi }_0)$. Twisting is invertible, so it is enough to prove nonsplitting for ${\chi }_0$.

Use the open Bruhat decomposition coordinateFor $x\ne 0$ the matrix factorizationshows that $g_f(x)=|x{|}^{-1}f(1)$ for all sufficiently large $|x|$: the lower-unipotent matrix approaches the identity, where $f$ is locally constant. The coordinate map is injective, since this open cell is dense in the projective-line quotient and induced functions are locally constant sections.

For $t=\mathrm{diag}(a,d)$, direct multiplication givesThere exists a section $f_0$ with $f_0(1)=1$ and coordinate functionIndeed the displayed factorization changes to the lower chart near infinity, where this tail corresponds to the constant function one. Thus it extends smoothly across the missing point. With $t=\mathrm{diag}(\varpi ,1)$ and $|\varpi |=q^{-1}$, the difference is the compactly supported functionNormalize additive Haar measure by $\mathrm{vol}(\mathfrak{o})=1$. ThenThis difference survives in the Jacquet module. To prove that assertion, the unipotent action translates coordinates: $g_{\pi (n(y))f}(x)=g_f(x+y)$. Any such translation difference is compactly supported because the tail is $f(1)|x{|}^{-1}$ and $|x+y|=|x|$ when $|x|$ is sufficiently large. Its integral is zero: integrate over a sufficiently large additive ball, invariant under translation by $y$, and change variables. A finite sum of unipotent translation differences also has integral zero. Hence $h$, whose integral is nonzero, cannot be a coinvariant relation. In particular, on $J$ the operator $\pi (t)-{\chi }_0(t)I$ is nonzero. This proves the equal-character Jacquet self-extension for GL2 is nonsplit.

Any $T$-equivariant functional $J\to {\chi }_0$ must kill the nonzero image of that operator. Since $J$ has dimension two, its space of such functionals has dimension at most one; the quotient in the supplied exact sequence already provides one. Twisting back gives the same conclusion for the common linear character $\chi =\lambda$. For any different $\xi$, a surjection $J\to \xi$ is impossible because both composition factors of $J$ are $\chi$. Combining the two cases givesWhen the two listed linear characters coincide, the dimension remains one; they are not two independent contributions.

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.

Fix a Borel subgroup $B$ and a maximal algebraic torus $T\subset B$. Conjugacy of Borel subgroups and the equality $N_G(B)=B$ identify the flag variety with $G/B$, by $gB\mapsto {}^{g}B$.

Every simultaneous conjugacy orbit on pairs contains a pair $(B,{}^{g}B)$: first conjugate its first member to $B$. The remaining allowed conjugations are by $B$. Two such pairs are in the same orbit exactly when $g^{\prime }$ belongs to $BgB$. Indeed ${}^{g^{\prime }}B={}^{bg}B$ is equivalent to $(bg{)}^{-1}{g}^{\prime }\in N_G(B)=B$. Thus there is an explicit bijectionThe Bruhat decomposition is the disjoint partition $G={\coprod }_{w\in W}B\dot{w}B$, with $W=N_G(T)/T$. Under the displayed bijection its double coset $B\dot{w}B$ becomes the orbitThis proves that the orbits are indexed by the Weyl group. Conversely, the assertion that these particular $O(w)$ exhaust the orbits without repetitions says precisely that the $B\dot{w}B$ exhaust $G$ without repetitions. Hence the orbit description and the Bruhat decomposition are equivalent, not merely two sets of the same cardinality. Their common index is the relative position of Borel subgroups.

For a rational Borel subgroup $B=TU$, the principal series is ${\mathrm{Ind}}_{B^F}^{G^F}\tilde{\theta }$, with $\tilde{\theta }$ inflated from a linear character of $T^F$. The Mackey restriction formula and rational Bruhat decomposition give its self-inner-product as the size of the Weyl stabilizer of $\theta$. It is irreducible precisely when that stabilizer is trivial.

The relative position of Borel subgroups $B_1,B_2$ is the Weyl group index of their simultaneous conjugacy orbit. After moving the first subgroup to a fixed $B$, the second has the form $gB{g}^{-1}$ and its index is the double coset $BgB$. This identifies the diagonal action on pairs of flags with the Bruhat decomposition.