4
$\begingroup$

Given a minimal parabloic subgroup we know that conjugation by the longest element in the weyl group takes it to the opposite parabolic.

Can we do the same thing if we choose a standard parabolic subgroup? Can we always find an element in the weyl group such that conjugation by this element takes it to the opposite parabolic? I am guessing this should be possible by modifying the longest element in the weyl group. Is this true? Any help is appreciated.

$\endgroup$
3
  • 3
    $\begingroup$ Please include a short gloss of your question in the title, so that folks reading only the front page know more about your question than merely the vague topic. "Can every parabolic subgroup be conjugated to its opposite by an element of the Weyl group?" is certainly short enough to fit (titles here can be about a tweet and a half in length). Otherwise, the question is a good one --- but if you feel like writing more, it's always nice to know why you're interested in this question, what you've already figured out, etc. $\endgroup$ Commented Apr 21, 2011 at 2:49
  • 1
    $\begingroup$ Adding to Theo's request, a tag algebraic-groups would be more appropriate here. The question (and negative answer) work uniformly for semisimple algebraic groups over arbitrary algebraically closed fields, by the way. Also, the terminology is out of focus. When a Borel subgroup $B$ is fixed and the rank is $r$, there are $2^r$ nonconjugate standard parabolics containing $B$. While $B$ itself is minimal in this collection, it's usual to call those with a single negative root "minimal"; similarly "maximal" includes "proper" in this context. $\endgroup$ Commented Apr 21, 2011 at 13:30
  • 2
    $\begingroup$ Is "a tweet" now really an official measurement of text-lengths? $\endgroup$ Commented Sep 2, 2011 at 14:20

3 Answers 3

8
$\begingroup$

If I understood your question correct, then the answer is no. I will assume for simplicity that you are talking about parabolic subgroups of complex simple Lie groups. Then your question translates to the corresponding question about closed subsystems of root systems. Recall that the standard parabolic subgroups bijectively correspond to the closed subsystems $R$ of the root system $\Phi$ such that $R$ contains the set of all positive roots $\Phi^+$. (Here `closed' means $\alpha\in R$, $\beta\in R$ implies $\alpha+\beta\in R$.) Your question is equivalent to the following one: given a closed subsystem $R$ containing all positive roots $\Phi^+$, is there an element $w$ of Weyl group such that $w.R=-R$?

If the Weyl group contains $-id$, then the question is yes. However, for the root systems of types $A_n$ ($n\ge2$), $D_{2n+1}$ and $E_6$ this is not true: $-id$ does not belong to the Weyl group. Therefore for these groups there may exist parabolic subgroups which are not conjugate to their opposites via the action of Weyl group. And indeed, looking at $A_2$, we see that the two standard maximal parabolics are not conjugate to their opposites via Weyl group: there is no element in Weyl group which transforms the root subsystem $R_1=\Phi^+\cup\{-\alpha_1\}$ to $-R_1$, and the same is true for $R_2=\Phi^+\cup\{-\alpha_2\}$.

$\endgroup$
5
$\begingroup$

I believe this is false, and that a counterexample occurs in $\mathrm{SL}(3)$ already.

Let $P$ be the block upper triangular subgroup with a $2\times2$ block in the top left corner. If I understand your terminology correctly, it's "opposite" is the block lower triangular group with the same block.

If a Weyl group element were to conjugate one to the other, it would have to map the root corresponding to the (1,2)-entry to that of the (2,1)-entry, and also (1,3) to (3,1). That is impossible for a single element of $S_3$.

$\endgroup$
1
  • 12
    $\begingroup$ In more down-to-earth terms, take a maximal parabolic subgroup of $SL(3)$. It acts naturally on a 3-dimensional vector space. It either stabilises a line but no plane, or a plane but no line, depending on which one you chose. Its opposite will do the opposite! So these parabolics cannot be conjugate in $GL(3)$. $\endgroup$ Commented Apr 21, 2011 at 7:45
-6
$\begingroup$

For real semisimple Lie groups with finite center the answer is YES: any two parabolic subgroups are conjugate by an element of the Weyl group.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .