1
$\begingroup$

Let $W$ be a Weyl group generated by the simple reflections $s_i$, $i \in I$, where $I$ is the vertex set of the Dynkin diagram of $W$. For $J \subset I$, let $W_J$ be the subgroup of $W$ generated by $s_j$, $j \in J$. Let $w_0^J$ be the longest word in $W_J$.

Suppose that $K \subset J \subset I$. I think that the following identity is true $w_0^J w_0^K = w_0^{K^*} w_0^J$, where ${}^*: J \to J$ is defined by $s_{j^*} = w_0^J s_j w_0^J$, $j \in J$.

Are there some reference about the proof of $w_0^J w_0^K = w_0^{K^*} w_0^J$? Thank you very much.

Improve this question
$\endgroup$
3
  • $\begingroup$ Is there any reason to work with $W$ and not $W_J$ as the ambient group? $\endgroup$
    – darij grinberg
    Commented Jun 14, 2018 at 20:44
  • 1
    $\begingroup$ Also, the proof of your claim should be a 5-liner if properly set up. Conjugation by $w_0^J$ is a group isomorphism $W_K \to W_{K^{\ast}}$ sending $K$ to $K^{\ast}$, so it is an isomorphism of Coxeter groups. Thus, it sends longest element to longest element. $\endgroup$
    – darij grinberg
    Commented Jun 14, 2018 at 21:14
  • $\begingroup$ @darij grinberg, thank you very much. Yes, we can take $W_J$ as the ambient group. $\endgroup$
    – Jianrong Li
    Commented Jun 15, 2018 at 9:39

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .