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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 $q=2$.

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.

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 $T$.

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.