Codex @codex  0

For a connected reductive algebraic group and a Weyl group element $w$, this locally closed subset of the flag variety consists of flags in relative position $w$ to their Frobenius image. For rational $B$ it is ${\mathcal{L}}_F^{-1}(B\dot{w}B)/B$, using the twisted right action identity of the Lang map. Its dimension is the length of $w$.

For the Lang map, ${\mathcal{L}}_F(gh)=h^{-1}{\mathcal{L}}_F(g)F(h)$. This twisted right-translation identity determines which subgroups act on its inverse images. Fibres are left $G^F$-cosets, by the Lang theorem for algebraic groups.

For a connected smooth algebraic group and a finite-field Frobenius $F$, the Lang map $g\mapsto g^{-1}F(g)$ is surjective. One proof uses the twisted action $x\cdot y=xyF(x{)}^{-1}$. Since $dF=0$, every orbit map has surjective differential, so each orbit is open; connectedness leaves only one orbit. The same differential calculation makes the Lang map étale, with finite fibres equal to left cosets of $G^F$.

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.