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For a connected reductive algebraic group, a Borel subgroup $B$, and a maximal torus $T\subseteq B$, the Weyl group $W=N_G(T)/T$ indexes the double cosets:

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 finite group of Lie type is obtained from rational points of a connected reductive algebraic group with a suitable Frobenius endomorphism of an algebraic group, often followed by central quotient or passage to a derived subgroup. Its complex representations combine finite-group character theory with the geometry of Borel subgroups, maximal algebraic tori, and the Weyl group.

For a connected reductive algebraic group $G$ and a Borel subgroup $B$, the quotient $G/B$ is its complete flag variety. Quotients $G/P$ by parabolic subgroups are partial flag varieties.

A maximal algebraic torus is an algebraic torus maximal among the tori contained in an algebraic group. In a connected reductive algebraic group, all maximal algebraic tori are conjugate. A Borel subgroup contains maximal algebraic tori, and its unipotent radical gives a decomposition $B=T⋉U$.

A parabolic subgroup of a connected reductive group is a closed subgroup containing a Borel subgroup. Its quotient in the group is projective.

Write $k={\overline{\mathbb{F}}}_q$, and let $F_0$ be the entrywise $q$th-power Frobenius endomorphism of an algebraic group on the general linear group $G={\mathrm{GL}}_2(k)$. The algebraic closure is visible in the original PDF; it is lost in the converted TeX. Let $T_0$ be the diagonal maximal algebraic torus and $B_0$ the upper triangular Borel subgroup.

Choose $\beta \in k$ with $\beta -{\beta }^q=1$. Such a $\beta$ exists because $k$ is an algebraic closure, and $\beta \notin {\mathbb{F}}_q$. PutTransporting the standard ${\mathbb{F}}_q$-model through the algebraic group automorphism $\mathrm{Int}(h_{\pm })$ gives a rational structure on an algebraic group over the same field ${\mathbb{F}}_q$. In particular these are genuine Frobenius endomorphisms of an algebraic group, and their fixed-point groups areThe auxiliary $\beta$ need not be rational for the original structure: it changes the identification with the geometric group, not the field over which the transported model is defined.

For either sign, take the rational maximal torus and Borel subgroupIndeed $F_{\pm }(h_{\pm }H{h}_{\pm }^{-1})=h_{\pm }{F}_0(H)h_{\pm }^{-1}$ for $H=T_0,B_0$, so both subgroups are $F_{\pm }$-stable. Conjugacy also preserves maximality of the algebraic torus and the property of being a Borel subgroup.

Finally,Neither $c_{+}$ nor $c_{-}$ is central, so neither endomorphism is $F_0$. If $F_{+}=F_{-}$, surjectivity of $F_0$ on $G(k)$ would imply that $c_{-}^{-1}{c}_{+}$ is central. Butis not scalar in any characteristic. Thus the two nonstandard rational structures have distinct Frobenius endomorphisms, including when $q=2$.

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.

Choose a rational Borel subgroup $B=TU$ with rational maximal algebraic torus $T$, and use the Lang map $\mathcal{L}(g)=g^{-1}F(g)$. Conjugating the pair $({}^{g}B,F({}^{g}B))$ by $g^{-1}$ gives $(B,{}^{\mathcal{L}(g)}B)$. Part (a) therefore givesMoreover $\mathcal{L}(gb)=b^{-1}\mathcal{L}(g)F(b)$, and $F(B)=B$, so this inverse image is stable under right multiplication by $B$. The flag variety identification consequently restricts to the required isomorphism

The second printed isomorphism is false as written: it omits a unipotent quotient. Take $G={\mathrm{SL}}_2({\overline{\mathbb{F}}}_q)$ with standard Frobenius endomorphism of an algebraic group and $w=1$. Then $X(1)$ consists of the $q+1$ rational lines in ${\mathbb{F}}_q^2$, so has dimension zero. Every Borel subgroup $B^{\prime }$ has one-dimensional unipotent radical $U^{\prime }$. The Lang map is a finite surjective étale morphism, so ${\mathcal{L}}^{-1}(U^{\prime })$ has dimension one. Quotienting by the finite group $T^{\prime F}$ preserves that dimension. Thus no choices of $T^{\prime }$ and $B^{\prime }$ can make the asserted two varieties isomorphic in this example.

Here is the corrected construction, including the missing quotient. Choose $x$ by the Lang theorem for algebraic groups so that $x^{-1}F(x)=\dot{w}$. DefineThen $F(T^{\prime })=x\dot{w}T{\dot{w}}^{-1}{x}^{-1}=T^{\prime }$, so $T^{\prime }$ is a rational maximal torus contained in $B^{\prime }$. Put $J=U_x\cap U^{\prime }=F^{-1}(U^{\prime })\cap U^{\prime }$. The correct unipotent quotient in a Deligne-Lusztig variety isThe semidirect product denotes the subgroup $J{T}^{\prime F}$ acting on the right. Both $U_x$ and $U^{\prime }$ are normalized by $T^{\prime }$, so the action is well defined. Alsofor $h\in {\mathcal{L}}^{-1}(U^{\prime })$, showing directly that the map lands in the Deligne-Lusztig variety.

To prove surjectivity and identify its fibres, use the intermediate varietyThe right $U$-action follows from the twisted-translation identity for the Lang map. Right multiplication by $T^{F_w}$ also acts, and $Y(\dot{w})/T^{F_w}\cong X(w)$. Indeed if $\mathcal{L}(g)\in B\dot{w}B$, write it as $u_{1}t\dot{w}{u}_2$. Replacing $g$ by $ga$, with $a\in T$, changes its torus factor to $a^{-1}t{F}_w(a)$. Surjectivity of the twisted Lang map on the connected algebraic torus makes this factor one. Once the factor is one, two representatives of the same flag differ by $ua$ with $u\in U$ and $a\in T^{F_w}$. This proves both surjectivity and the asserted finite covering, including its group of fibres.

Now map $Z={\mathcal{L}}^{-1}(U^{\prime })$ to $Y(\dot{w})$ by $h\mapsto hxU$. Given $\mathcal{L}(g)=u_1\dot{w}{u}_2$, replace $g$ by $g{u}_1$. ThenThus $h=g{u}_1{x}^{-1}$ belongs to $Z$ and maps to the original point $gU$. This proves surjectivity. Two preimages have the form $h$ and $hk$, where $k\in U_x$. Since $F(k)\in U^{\prime }$ and $\mathcal{L}(h)\in U^{\prime }$, the identityshows that $hk\in Z$ precisely when $k\in U^{\prime }$, hence precisely when $k\in J$. Therefore the fibres are exactly the right $J$-orbits.

These are isomorphisms of geometric quotients, not just bijections on points. The groups $U_x,U^{\prime }$ contain root subgroups for the same maximal algebraic torus, so their intersection $J$ is smooth. The map $Z\to Y(\dot{w})$ is a smooth surjection with fibre $J$: the differential of the Lang map is invertible, and the kernel of the differential of $h\mapsto hxU$ on $Z$ is the translate of $\mathrm{Lie}(U_x)\cap \mathrm{Lie}(U^{\prime })=\mathrm{Lie}(J)$. This gives $Z/J\cong Y(\dot{w})$. Finally $x{T}^{F_w}{x}^{-1}=T^{\prime F}$, so quotienting the remaining finite torus action gives the displayed corrected formula.

For $w=1$, $J=U_x$ is exactly the affine-line factor in the counterexample. More generally $Z/T^{\prime F}\to X(w)$ retains unipotent fibres of dimension $\dim J$; it becomes the printed isomorphism only when this intersection is trivial. The missing factor is also recorded in the quotient construction in Tiep's lectures, proof of Theorem 5.22.

For $G={\mathrm{SL}}_2$ and the upper triangular Borel subgroup $B$, the quotient is $G/B\cong {\mathbb{P}}^1$: a matrix is sent to the line spanned by its first column. Over $U_0=\{[1:t]\}$ use the sectionand over $U_1=\{[s:1]\}$ useEvery matrix above $U_i$ is uniquely $s_i(u)b$ with $b\in B$. Thus each inverse image is $U_i\times B$, proving that $G\to G/B$ is a Zariski $B$-torsor.

Embed the space $\mathcal{B}$ of complete isotropic flags intoThe incidence conditions $F_i\subset F_{i+1}$ and isotropy equations $⟨F_i,F_i⟩=0$ are closed polynomial conditions. Since the Grassmannian is projective, $\mathcal{B}$ is a projective algebraic variety.

Every complete isotropic flag extends to a symplectic basis. A symplectic change of basis carries any such flag to any other, so $G={\mathrm{Sp}}_{2n}$ acts transitively. The stabilizer $B$ of the standard flag consists of the upper triangular symplectic matrices. It is closed, connected, and solvable. The Lie-Kolchin theorem shows that every connected solvable subgroup fixes a complete flag in $V$; preservation of the symplectic form makes the resulting flag isotropic after taking its first half. Such a subgroup is conjugate into $B$, so $B$ is maximal and hence a Borel subgroup.

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.

A maximal algebraic torus $T$ is rational for $F$ when $F(T)=T$. Rationality does not require every point of $T$ to be fixed: the finite group $T^F$ is its rational-point group. Similarly a rational Borel subgroup or parabolic subgroup means an $F$-stable algebraic subgroup.