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Some Notions

  • A variety over a field is defined to be a scheme of finite type over this field.

  • An $\mathbb{F}_q$-rational structure of an $\bar{\mathbb{F}}_q$-variety $V$ is a $\mathbb{F}_q$-variety $V_0$ with an isomorphism $V\cong V_0\times_{\mathbb{F}_q}\bar{\mathbb{F}}_q$ of varieties over $\bar{\mathbb{F}}_q$.

  • Along with the above notations, let $\mathrm{Frob}$ be the absolute Frobenius endomorphism of $V_0$, and we call $F:=\mathrm{Frob}\otimes \mathrm{ Id_{\bar{\mathbb{F}}_q}}$ the geometric Frobenius endomorphism of $V$ associated to the rational structure given by $V_0$. It is clear that geometric Frobenius endomorphisms and rational structures are $1-1$ corresponded.

  • Along with the above notations, we call a subvariety $W$ of $V$ is rational if $FW=W$. In this case one can show $W$ also admits a rational structure by the restriction of $F$.

  • More infomation about these terminologies can be found in Digne & Michel Representations of Finite Groups of Lie type, and Malle & Testerman Linear Algebraic Groups and Finite Groups of Lie Type.

An Example

Different geometric Frobenius morphisms can be very different. Consider the general linear group $GL_n(\bar{\mathbb{F}}_q)$, then it has a natural geometric Frobenius $F$ (the one taking each entry of a matrix to its $q$-th power) with fixed points $$GL_n(\bar{\mathbb{F}}_q)^F=GL_n(\mathbb{F}_q).$$ Meanwhile, $F'(-):=(F(-)^T)^{-1}$, where $T$ means taking transpose, defines another geometric Frobenius (hence another rational structure over $\mathbb{F}_q$), and it has fixed points $$GL_n(\bar{\mathbb{F}}_q)^{F'}=U_n(\mathbb{F}_q)$$ the unitary group over $\mathbb{F}_q$.

My Question

From now on let $G$ be a connected reductive group over $\bar{\mathbb{F}}_q$, with a rational structure over $\mathbb{F}_q$, denote by $F$ the associated geometric Frobenius endomorphism, and let $B$ be a rational Borel subgroup. Then $F$ acts on the closed points of the flag variety $X=G/B$ over $\bar{\mathbb{F}}_q$.

My Question: Does the action of $F$ on $X$ give an $\mathbb{F}_q$-rational structure on $X$, i.e. does $F$ act on $X$ as a geometric Frobenius with respect to some rational structure?

Any references or suggestions are appreciated.


Some Ideas

I'm not familiar with the construction of quotients, but I am trying to split the question into the following (possibly vague) steps, please ignore them if they are just don't make much sense:

  1. Can the action of $F$ on $X$ be realized as a morphism of schemes (Carnahan confirmed $F$ is a morphism in the below comment, thanks)?

  2. Let $V_0$ be the rational structure of $G$ associated to $F$, and let $W_0$ be the ratoinal structure of $B$, must $V_0$ be a group scheme over $\mathbb{F}_q$, is it always the case that $W_0$ be a closed subscheme of $V_0$?

  3. Does the quotient scheme $V_0/W_0$ exist?

  4. Should we have that $V_0/W_0\times_{\mathbb{F}_q}\bar{\mathbb{F}}_q\cong G/B$, and $F=\mathrm{Frob}_{V_0/W_0}\otimes \mathrm{Id}_{\bar{\mathbb{F}}_q}$?

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  • $\begingroup$ What do you mean by a $\mathbb{F}_q$-rational structure on $X$? Do you want to know if $X$ is defined over $\mathbb{F}_q$? $\endgroup$ Jul 20, 2014 at 17:03
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    $\begingroup$ What do you mean by "does $F$ give the rational structure"? If $G$ and $B$ are both defined over $\mathbb{F}_q$, then so is $X$, by construction. This has nothing to do with Frobenius elements. $\endgroup$ Jul 20, 2014 at 18:46
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    $\begingroup$ It is not the action of $F$ that gives $X$ the rational structure. For any field $k$ (or indeed any base scheme), and any affine $k$-group $G_0$, the quotient by a closed affine subgroup $B_0$ is a $k$-scheme. In your case, $F$ acts on $G/B$ by $\mathbb{F}_q$-morphisms. $\endgroup$
    – S. Carnahan
    Jul 23, 2014 at 9:21
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    $\begingroup$ I do not understand why this question was put on hold as it is perfectly clear what is being asked in the context of common terminology used for algebraic groups over finite fields; see for instance Digne & Michel, Representations of Finite Groups of Lie type, or Malle & Testerman, Linear Algebraic Groups and Finite Groups of Lie Type. $\endgroup$ Jul 24, 2014 at 10:38
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    $\begingroup$ @user148212: On your variety $X=G/B$ you already have a canonical structure of $G$-variety (over $\overline{F}_q$) and a canonical rational point $eB=B$, the image of the unit element $e$ of $G$. Therefore, the $F_q$-rational structure on $X$ is unique, if exists. What is not evident is the existence of an $F_q$-rational structure. This can be proved by an explicit construction of a closed (in your case) subvariety in a projective space, see the proof of Corollary 5.5.4 in Springer's book mentioned by Daniel Loughran. $\endgroup$ Jul 24, 2014 at 21:39

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