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Take $G=\operatorname{GL}_3$, defined over the algebraic closure of a finite field $\mathbb{F}_q$ and let $X$ be the set of Borel subgroups of $G$. The Frobenius morphism $F:G\to G$ induces a map $F:X\to X$, as $G/B\cong X$ for any chosen Borel subset $B$. The Weyl group $W$ of $G$ is isomorphic to the symmetric group $S_3$, and we have a bijection between $W$ and the set of $G$-orbits on $X\times X$. Fix a Borel subset $B^+\in X$. For $w\in W$, define $\mathcal{O}(w)$ to be the orbit of $(B^+,\dot wB^+\dot w^{-1})$ in $X\times X$. We say that $B_1,B_2\in X$ are in relative position $w$ if $(B_1,B_2)\in\mathcal{O}(w)$.

For any $w\in W$, we can define the Deligne–Lusztig variety associated to it as $X(w)=\{B\in X:B\text{ and }F(B)\text{ are in relative position }w\}$.

My Question

I want to compute $X((1\ 2))$.

I know that $X$ can be identified with the full flag variety $\mathcal{F}_3$, but how I can interpret the condition of $B$ and $F(B)$ being in relative position $w$ in terms of flags?

Any help is appreciated.

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  • $\begingroup$ Terminology: you say that you want to take $G = \operatorname{GL}_{3\,\overline{\mathbb F_q}}$; but you actually want $G = \operatorname{GL}_{3\,\mathbb F_q}$, or else there is no canonical Frobenius endomorphism of $G_{\overline{\mathbb F_q}}$. $\endgroup$
    – LSpice
    Commented May 18, 2023 at 16:56
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    $\begingroup$ Thanks! I think what I wanted to say is that I take $G_0=\text{GL}_{3,\mathbb{F}_q}$ and then $G=G_0\times_{\mathbb{F}_q}\overline{\mathbb{F}_q}$, the base change to the algebraic closure. Do you agree? $\endgroup$
    – EJB
    Commented May 19, 2023 at 8:50
  • $\begingroup$ Re, yes, I agree. (I meant $G_0 \times_{\mathbb F_q} \overline{\mathbb F_q}$ by $(G_0)_{\overline{\mathbb F_q}}$.) $\endgroup$
    – LSpice
    Commented May 19, 2023 at 14:12

1 Answer 1

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Let $\mathcal{F}\colon V_1\subseteq\dotsb\subseteq V_n$ and $\mathcal{F}'\colon U_1\subseteq\dotsb\subseteq U_n$ be two complete flags. Then $\mathcal{F}$ and $\mathcal{F}'$ are said to be in relative position $w$, where $w\in S_n$ is a permutation, if $$\dim V_i\cap U_j=\#(\{ 1,\dotsc,i \}\cap \{ w(1),\dots,w(j) \})$$ for any $i,j\in\{1,\dotsc,n\}$.

In particular, the variety $X((1\ 2))$ consists of the flags $V_1\subseteq V_2$ in which $V_2$ is an $F$-stable plane and $V_1$ is an $F$-unstable line. So it is $$\operatorname{Gr}(1,3)^F\times \mathbb{P}^1\setminus\mathbb{P}^1(\mathbb{F}_q),$$ a disjoint union of copies of the Drinfeld curve.

Computations like this are very useful for making explicit connections with combinatorics. For a reference, you may want to take a look at Steinberg's paper An occurrence of the Robinson–Schensted correspondence.

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  • $\begingroup$ Yes, it means the complement. $\endgroup$
    – user148212
    Commented May 18, 2023 at 17:21
  • $\begingroup$ Re, OK, I edited accordingly. $\endgroup$
    – LSpice
    Commented May 18, 2023 at 18:11
  • $\begingroup$ Many many thanks!! This is exactly what I need. I guess that I now need to show why this works, i.e. that a Borel subgroup $B$ is in relative position $w$ with $F(B)$ exactly if the flags are in relative position $w$. Do you have any hints? $\endgroup$
    – EJB
    Commented May 19, 2023 at 8:53
  • $\begingroup$ You may first transfer to the standard upper triangular Borel, then take a standard basis on which the permutations permute. $\endgroup$
    – user148212
    Commented May 19, 2023 at 14:24
  • $\begingroup$ What do you mean exactly with "transfer to the standard upper triangular Borel"? Suppose we denote this Borel subgroup by $B^+$, and another Borel $B$ is in relative position $w$ with $F(B)$. Do you mean that there is a $g\in G$ such that $B=gB^+g^{-1}$ and $F(B)=g\dot w B^+\dot w^{-1}g^{-1}$? Here, I am using the orbit definition of relative position. $\endgroup$
    – EJB
    Commented May 19, 2023 at 14:57

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