0
$\begingroup$

Let $\Phi$ be a root system and $\gamma \in \Phi$ a root. Let $W$ be the Weyl group and $\Delta$ a set of simple roots. Let $w \in W$ such that $w(\gamma)=\gamma$. Is it true that if $w=s_1\dots s_n$ with $s_i$ simple reflections, then $s_i(\gamma)=\gamma$ for $i=1\dots n$?

I can't see a reason why this shouldn't be true, but am unable to prove it.

Thanks!

$\endgroup$
4
$\begingroup$

This is definitely not true. For instance, already in $\Phi=B_2$, each root has a root orthogonal to it, so for every root there is some nontrivial element (in fact, a reflection) of the Weyl group fixing it. But e.g. a simple root in $B_2$ is not fixed by any simple reflection.

However, what you might want to know is the following. If we choose any point $v$ in the dominant chamber of our root system, then the stabilizer of $v$ is exactly the parabolic subgroup of $W$ generated by simple reflections that fix $v$. For a proof of this see Lemma 10.3B of Humphreys' "Introduction to Lie Algebras and Representation Theory" (https://www.springer.com/us/book/9780387900537).

$\endgroup$

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.