Timeline for When the longest element of Weyl group is rational?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Apr 11, 2017 at 20:54 | vote | accept | user148212 | ||
| Apr 11, 2017 at 13:33 | answer | added | Jay Taylor | timeline score: 8 | |
| Apr 11, 2017 at 13:31 | history | edited | user148212 | CC BY-SA 3.0 |
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| Apr 11, 2017 at 13:27 | comment | added | user148212 | @JasonStarr You are right, it should be defined with respect to a Borel $B$. The longest element is then the element corresponding to the open piece in $G=\cup_w BwB$. But still, if $B$ is not $F$-rational, the longest element may not be $F$-rational. I will edit my question. | |
| Apr 11, 2017 at 13:14 | comment | added | Jason Starr | 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"? | |
| Apr 11, 2017 at 13:06 | comment | added | Jason Starr | . . . The natural projection $\text{pr}_2:\mathcal{B}\times_{\text{Spec}(k)}\mathcal{B}\to \mathcal{B}$ is $G$-equivariant. Thus, every orbit surjects under $\text{pr}_2$. The fiber over a geometric point $B$ is simply a copy of $G/B$ with its induced $B$-action. The orbits are the Bruhat cells $BwB/B$. The inverse images of these orbits in $G\times_{\text{Spec}(k)} \mathcal{B}$ are the usual Bruhat cells. | |
| Apr 11, 2017 at 13:04 | comment | added | user148212 | @JasonStarr I am a bit confused. Does the rationality of the maximal stratum implies the rationality of the Weyl element? | |
| Apr 11, 2017 at 13:04 | comment | added | Jason Starr | Your Bruhat decomposition is a special case of the Bruhat decomposition of $\mathcal{B}\times_{\text{Spec}(k)}\mathcal{B}$ (where $k$ is the field over which $G$ is defined). To make this more clear, consider the natural smooth surjective morphism $G\times_{\text{Spec}(k)}\mathcal{B} \to \mathcal{B}\times_{\text{Spec}(k)} \mathcal{B}$ given (on the level of functors of points) by $(g,B)\mapsto (gBg^{-1},B)$. The Bruhat decomposition of $\mathcal{B}\times_{\text{Spec}(k)}\mathcal{B}$ is simply the orbit decomposition for the natural diagonal action of $G$ . . . | |
| Apr 11, 2017 at 12:50 | comment | added | Jason Starr | 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. | |
| Apr 11, 2017 at 12:43 | comment | added | user148212 | @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. | |
| Apr 11, 2017 at 12:40 | comment | added | user148212 | @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. | |
| Apr 11, 2017 at 12:18 | comment | added | Jason Starr | 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 $W(T)$? | |
| Apr 11, 2017 at 12:03 | history | asked | user148212 | CC BY-SA 3.0 |