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Let $\mathbb{G}$ be a connected reductive group over $\mathbb{F}_q$, and let $G$ be the base change to an algebraic closure of the base field. Denote by $F$ the associated geometric Frobenius.

Let $T$ be an $F$-rational (i.e. $FT=T$) maximal torus of $G$. As $T$ is $F$-rational, the Frobenius $F$ acts on the finite Weyl group $W(T)$. In general, the subgroup $W(T)^F$ consisting of $F$-fixed points is a proper subgroup, and depends on the choice of $T$.

How to determine when the longest element of $W(T)$ appears in $W(T)^F$? Is there a (maybe case by case) list available?

As commented below, the longest element is with respect to a Borel subgroup $B$ containing $T$: Then the longest element is the element corresponding to the open piece in $G=\cup_{w\in W(T)}BwB$.

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    $\begingroup$ Are you specifying an $F$-rational Borel that contains $T$? If not, how are you defining the simple reflections, resp. the length of elements, in $W(T)$? $\endgroup$
    – Jason Starr
    Commented Apr 11, 2017 at 12:18
  • $\begingroup$ @JasonStarr No, I do not assume T is contained in an $F$-rational Borel. Roots can be defined by considering the conjugation actions of $T$ on minimal unipotent subgroups; when you specified the positive roots (i.e. choose a Borel), the simple roots are the indecomposable roots. These can be found e.g. in 0.26 and 0.31 of Digne--Michel's Representations of Finite Groups of Lie Type. $\endgroup$
    – user148212
    Commented Apr 11, 2017 at 12:40
  • $\begingroup$ @JasonStarr And if I understand it correctly, the longest element can be defined without the term of roots: It is the element corresponding to the maximal strata in the Bruhat decomposition. $\endgroup$
    – user148212
    Commented Apr 11, 2017 at 12:43
  • $\begingroup$ Are you defining the Bruhat decomposition of $\mathcal{B}\times_{\text{Spec}(F)}\mathcal{B}$, where $\mathcal{B}$ is the $G$-homogeneous scheme over $\text{Spec}(F)$ representing the functor of Borel subgroups? In that case, the maximal stratum can be defined as the open subscheme parameterizing those pairs of Borels whose intersection is a maximal torus. So the maximal stratum is $F$-rational. $\endgroup$
    – Jason Starr
    Commented Apr 11, 2017 at 12:50
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    $\begingroup$ I guess that is what I was saying, but now I have a concern. If $w$ is an element of the Weyl group such that $wBw^{-1}$ and $B$ are opposite Borels, then for every element $s$ of the Weyl group, for the Borel $B'=sBs^{-1}$, it seems that the opposite Borel to $B'$ is obtained by conjugating by $sws^{-1}$, not by $w$. So now I renew my original question: how are you defining an element of "longest length"? $\endgroup$
    – Jason Starr
    Commented Apr 11, 2017 at 13:14

1 Answer 1

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Let $B$ be a Borel subgroup containing $T$. As $F(B)$ and $B$ are both Borel subgroups containing $T$ there exists an element $n \in N_G(T)$ such that ${}^nF(B) = B$. Thus the Frobenius endomorphism $F' : G \to G$ defined by $F'(g) = nF(g)n^{-1}$ induces an automorphism $F' : W \to W$, where $W = N_G(T)/T$, and this stabilises the Coxeter generators $\mathbb{S} \subseteq W$ determined by $B$. Thus $F'$ is a length preserving automorphism so must fix the longest element (by uniqueness). Hence, $F$ fixes the longest element if and only if $\bar{n} \in C_W(w_0)$, where $\bar{n} \in W$ is the image of $n$.

One can ask how unique the element $n$ is. Assume $m \in N_G(T)$ also satisfies ${}^mF(B) = B$ then $mn^{-1} \in N_G(B) = B$ so $mn^{-1} \in N_B(T) = T$ thus $\bar{n} = \bar{m}$. Hence the condition that $F$ fixes the longest element does not depend upon the choice of element $n$.

Let's see how to compute this in practice. Choose an $F$-stable maximal torus and Borel subgroup $T_0 \leqslant B_0 \leqslant G$. There then exists an element $x \in G$ such that $(T,B) = ({}^xT_0,{}^xB_0)$. Note, by the above argument that such an element $x$ is unique up to right multiplication by elements of $T_0$. Let $(W_0,\mathbb{S}_0)$ and $(W,\mathbb{S})$ be the Coxeter system defined with respect to $(T_0,B_0)$ and $(T,B)$ respectively. Then conjugation by $x$ induces an isomorphism $W_0 \to W$ mapping $\mathbb{S}_0$ onto $\mathbb{S}$. Hence, if $w_0 \in W_0$ is the longest element then ${}^xw_0 \in W$ is the longest element.

Assume $x^{-1}F(x) = n \in N_G(T_0)$ then we have $$F(B) = F({}^xB_0) = {}^{xn}B_0 = {}^{xnx^{-1}}B.$$ Hence $xn^{-1}x^{-1} \in N_G(T) = {}^xN_G(T_0)$ is an element as above. We have $x\bar{n}^{-1}x^{-1} \in C_W({}^xw_0) = {}^xC_{W_0}(w_0)$ if and only if $\bar{n} \in C_{W_0}(w_0)$. Thus one can easily construct examples where the longest element is not fixed.

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  • $\begingroup$ Thank you. Does this property actually only depends on $T$, i.e. is it possible that there are $x$ and $y$ such that ${^xT_0}={^yT_0}$ but $x^{-1}F(x)$ represents an element in the centraliser while $y^{-1}F(y)$ does not? $\endgroup$
    – user148212
    Commented Apr 11, 2017 at 15:53
  • $\begingroup$ Yes, one can already see this from the first part of the answer. I'll update the answer to make it clearer. $\endgroup$
    – Jay Taylor
    Commented Apr 11, 2017 at 19:54
  • $\begingroup$ The point here is that $x$ is unique up to right multiplication by elements of $T_0$. $\endgroup$
    – Jay Taylor
    Commented Apr 11, 2017 at 20:26

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