Let $W$ be the Weyl group of a root system $\Phi$ with base $\Delta$ and system of positive roots $\Phi^+$. Let $S = \{ w_{\alpha} : \alpha \in \Delta \}$ be the set of simple reflections corresponding to elements of $\Delta$. Let $\theta \subset \Delta$, and let $w_0 = w_l w_{l,\theta}$, where $w_l$ and $w_{l,\theta}$ are the long elements of $W$ and $W_{\theta} = \langle w_{\alpha} : \alpha \in \theta \}$.

The element $w_0$ is characterized uniquely by the following property: in each left coset of $W_{\theta}$ in $W$, there is a unique representative of smallest length, and $w_0$ is the longest of all these special representatives. It has the property that $\ell(w_0x) = \ell(w_0) + \ell(x)$ for all $x \in W_{\theta}$.

I noticed it follows immediately from here that if $x \in W_{\theta}$, then $w_0x \geq w_0$ in the Bruhat order. I was wondering whether the converse is true. So my question is:

If $w \in W$, and $w \geq w_0$ in the Bruhat order, then do we have $w = w_0x$ for some $x \in W_{\theta}$?

  • 2
    $\begingroup$ Note that Nathan's answer implies that your assumption about $W$ being a Weyl group is too restrictive; it just needs to be a finite Coxeter group. $\endgroup$ Sep 16, 2019 at 19:07

1 Answer 1


Yes, this follows from the fact that $x \mapsto w_l x$ is an antiautomorphism of the Bruhat order on a finite Coxeter group. (See Björner and Brenti, Proposition 2.3.4, for example, but their $w_0$ is your $w_l$) You also need the fact that $w_l$ is an involution. (For example, Björner and Brenti, Proposition 2.3.2.)

If $w\geq w_0$, then $w_lw\leq w_{l,\theta}$, which further implies that $w_lw\in W_\theta$. Thus $w=w_l(w_lw)=w_0(w_{l,\theta}w_lw)$, so the desired $x$ is $w_{l,\theta}w_lw$.


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.