Let $G$ be a connected reductive group defined over a finite field $F_q$ of characteristic $p >0$ with $q$ elements and corresponding Frobenius map $F : G → G$. Let $T$ be an $F$- stable maximal torus of $G$ contained in an $F$-stable Borel subgroup $B$ of $G$, and let $Φ$ be the root system of $G$ relative to $T$.

For $α ∈ Φ$, we write $X_α$ for the unipotent subgroup of $G$ corresponding to $α$. That is, $X_α$ is the minimal non-trivial closed unipotent subgroup of $G$ normalized by $T$, such that $T$ acts on $X_α$ by $α$, where $α$ is viewed as a character of $T$. Recall that $X_α$ and $(\overline{F_p},+)$ are isomorphic as algebraic groups. Fix an automorphism $x_α : F_p →X_α$ for every $α ∈ Φ$.

Assume that $F' : G→G$ be a Frobenius map of $G$ commuting with $F$ such that $T$ and $B$ are $F'$-stable.

We say that $F'$ has property (*) if for every $α ∈ Φ$, there is a non-negative integer $m > 0$ such that for all $t ∈ \overline{F_p}$ we have $F'(x_α(t)) = x_{F' (α)}(t^{p^m}) $.

My Question Let's consider $G=SL_n(q)$, and $F': G \rightarrow G$ be a field automorphism i.e., $F'((a_{ij}))=(a_{ij}^{p^k})$ for some $k\in N$. Does $F'$ satisfy the property (*)?

My Thoughts:

By a result of Steinberg [Endomorphisms of linear algebraic groups, Section 11.2], $F'$ permutes the roots; in fact, $F'$ permutes the root subgroups by $F'(X_{\alpha})=X_{{\alpha}'}$ and this gives rise to a permutation $b:\Phi \rightarrow \Phi$, in which $b(\alpha):={\alpha}'$. Moreover, for each $\alpha \in \Phi$, we have $F'(x_{α}(t)) = x_{b(α)}(d_{\alpha} t^{q_{\alpha}})$ for some $d_\alpha \in \overline{F_p}^*$ and $t\in \overline{F_p}^+$.

On the other hand, for groups of type $A_n$, the root subgroups $X_{\alpha}$'s are all of the form $\lbrace I+ kE_{ij} : k \in \overline{F_p}^+ \rbrace$ for $1\leq i, j \leq n$ (where $I$ is identity matrix and $E_{ij}$ is a matrix which is zero everywhere except for the entry $ij$th). Therefore, according to the definition of $F'$, we have $F'(X_{\alpha})=X_{{\alpha}}$ for all $\alpha$, and this means that the permutation $b$ acts trivially on roots. Hence $F'(x_{α}(t)) = x_{α}(d_{\alpha} t^{q_{\alpha}})$. So I am done if I could show that $d_{\alpha}=1$ for all $\alpha$. Is there any way to finish this argument?


An isomorphism $x_\alpha: \overline{\mathbb{F}}_p \to X_\alpha$ can always be composed with an automorphism of $\overline{\mathbb{F}_p}$ to give another such isomorphism. Note that $F'(x_\alpha(d t)) = x_\alpha(d_\alpha d^{q_\alpha} t^{q_\alpha})$, so you can just choose $d \in \overline{\mathbb{F}_p}^*$ such that $d_\alpha d^{q_\alpha} = d$ and replace $x_\alpha$ by $t \mapsto x_\alpha(d t)$ to obtain an $x_\alpha$ for which $F'$ satisfies property (*).

However, $F'$ will not have property (*) for arbitrary choices of $x_\alpha$.

  • $\begingroup$ I think you meant $d_{\alpha} d^{q_{\alpha}}=1$...Anyway, thank you so much for the answer. $\endgroup$ – user97635 May 16 '17 at 10:16
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    $\begingroup$ @user97635 : Actually, no, I mean $d_\alpha d^{q_\alpha} = d$. You want to have $F(x_\alpha(d t)) = x_\alpha(d t^{q_\alpha})$ and since $F(x_\alpha(d t)) = x_\alpha(d_\alpha d^{q_\alpha} t^{q_\alpha})$, you need to have $d_\alpha d^{q_\alpha} t^{q_\alpha} = d t^{q_\alpha}$ for all $t$, hence $d_\alpha d^{q_\alpha} = d$. $\endgroup$ – Matthias Klupsch May 17 '17 at 6:14

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