5
$\begingroup$

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 $G=\cup_{w\in W(T)}BwB$.

$\endgroup$
  • 1
    $\begingroup$ 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 Apr 11 '17 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 Apr 11 '17 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 Apr 11 '17 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 Apr 11 '17 at 12:50
  • 2
    $\begingroup$ 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 Apr 11 '17 at 13:14
8
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Apr 11 '17 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 Apr 11 '17 at 19:54
  • $\begingroup$ The point here is that $x$ is unique up to right multiplication by elements of $T_0$. $\endgroup$ – Jay Taylor Apr 11 '17 at 20:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.